代码笔记_图论

作者： 作者的头像   zqiceberg ,  2020-01-17 07:00:38 ,  所有人可见 ,  阅读 1225

-2


5

树与图的深度优先遍历

AcWing 846. 树的重心

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1e5 + 10, M = N * 2;

int n;
int h[N], e[M], ne[M], idx; //数组开错了，TLE
bool st[N];
int ans = N;

void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

int dfs(int u)
{
    st[u] = true;

    int sum = 1, res = 0;
    for (int i = h[u]; i != -1; i = ne[i])
    {
        int j = e[i];
        if (!st[j])
        {
            int s = dfs(j);
            res = max(res, s); //所有子树的最大值
            sum += s;          //累积求这个点的子树大小
        }
    }

    res = max(res, n - sum);   //这个点上面的那个连通块，比一下大小
    ans = min(ans, res);       //题目求的是最大值最小

    return sum;  //返回u点子树的大小
}

int main()
{
    cin >> n;
    memset(h, -1, sizeof h);

    for (int i = 0; i < n - 1; i++)
    {
        int a, b;
        //cin >> a >> b;
        scanf("%d%d", &a, &b);
        add(a, b), add(b, a);
    }

    dfs(1);

    cout << ans << endl;

    return 0;
}

作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/116289/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

树与图的广度优先遍历

AcWing 847. 图中点的层次

#include <iostream>
#include <cstring>

using namespace std;

const int N = 1e5 + 10;

int n, m;
int h[N], e[N], ne[N], idx;
int d[N], q[N];

void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

int bfs()
{
    int hh = 0, tt = 0;
    q[0] = 1;

    memset(d, -1, sizeof d);

    d[1] = 0;
    while (hh <= tt)
    {
        int t = q[hh++];
        for (int i = h[t]; i != -1; i = ne[i])
        {
            int j = e[i];
            if (d[j] == -1)
            {
                d[j] = d[t] + 1;
                q[++tt] = j;
            }
        }

    }

    return d[n];
}

int main()
{
    cin >> n >> m;
    memset(h, -1, sizeof h);

    for (int i = 0; i < m; i++)
    {
        int a, b;
        cin >> a >> b;
        add(a, b);
    }

    cout << bfs() << endl;

    return 0;
}


作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/116318/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

拓扑排序

AcWing 848. 有向图的拓扑序列

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int N = 1e5 + 10;

int n, m;
int h[N], e[N], ne[N], idx;
int d[N], q[N];

void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

bool topsort()
{
    int hh = 0, tt = -1;
    for (int i = 1; i <= n; i++)
        if (!d[i]) q[++tt] = i;

    while (hh <= tt)
    {
        int t = q[hh++];
        for (int i = h[t]; i != -1; i = ne[i]) //枚举t的所有出边t -> j
        {
            int j = e[i];  //出边
            d[j] --;       //干掉这条边，就是j的入度-1
            if (d[j] == 0) q[++tt] = j;
        }
    }

    if (tt == n - 1) return true;
    else return false;
}

int main()
{
    scanf("%d%d", &n, &m);
    memset(h, -1, sizeof h);

    for (int i = 0; i < m; i++)
    {
        int a, b;
        scanf("%d%d", &a, &b);

        add(a, b);
        d[b]++;  //b的入度+1
    }

    if (topsort())
    {
        for (int i = 0; i < n; i++)
            printf("%d ", q[i]);
    }
    else
        puts("-1");

    return 0;
}

作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/116336/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

Dijkstra

AcWing 849. Dijkstra求最短路 I

#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 510;

int n, m;
int g[N][N];
int dist[N];
bool st[N];

int dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;

    for (int i = 0; i < n; i++)
    {
        int t = -1;
        for (int j = 1; j <= n; j++)
            if (!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;

        st[t] = true;

        for (int j = 1; j <= n; j++)
            dist[j] = min(dist[j], dist[t] + g[t][j]);
    }

    if (dist[n] == 0x3f3f3f3f) return -1;  //这个地方== 被打成了=，调试了半天
    else dist[n];
}

int main()
{
    scanf("%d%d", &n, &m);

//    for (int i = 1; i <= n; i++)
//       for (int j = 1; j <=n; j++)
//            if (i == j) g[i][j] = 0;
//            else g[i][j]= INF;
    memset(g, 0x3f, sizeof g);

    while (m--)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        g[a][b] = min(g[a][b], c);  //重边，邻接矩阵中只存最小的权值
    }

    int t = dijkstra();

    printf("%d\n", t);

    return 0;
}

作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/123129/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

AcWing 850. Dijkstra求最短路 II

#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>

using namespace std;

typedef pair<int, int> PII;

const int N = 100010;

int n, m;
int h[N], w[N], e[N], ne[N], idx;
int dist[N];
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

int dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;

    priority_queue<PII, vector<PII>, greater<PII>> heap;
    heap.push({0, 1});

    while (heap.size())
    {
        auto t = heap.top();
        heap.pop();

        int ver = t.second, distance = t.first;
        if (st[ver]) continue;

        for (int i = h[ver]; i != -1; i = ne[i])
        {
            int j = e[i];
            if (dist[j] > distance + w[i])
            {
                dist[j] = distance + w[i];
                heap.push({dist[j], j});
            }
        }
    }

    if (dist[n] == 0x3f3f3f3f) return -1; //==, = 再一次打错
    return dist[n];
}

int main()
{
    scanf("%d%d", &n, &m);

    memset(h, -1, sizeof h);

    while (m--)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c);
    }

    int t = dijkstra(); 

    printf("%d\n", t);

    return 0;
}

作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/123562/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

Bellman-ford

AcWing 853. 有边数限制的最短路

#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 510, M = 10010;

int n, m, k;
int dist[N], backup[N];

struct Edge
{
    int a, b, w;
}edges[M];

int bellman_ford()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;

    for (int i = 0; i < k; i++)
    {
        memcpy(backup, dist, sizeof dist);
        for (int j = 0; j < m; j++)
        {
            int a = edges[j].a, b = edges[j].b, w = edges[j].w;
            dist[b] = min(dist[b], backup[a] + w);
        }
    }

    if (dist[n] > 0x3f3f3f3f / 2) return -1;
    //if (dist[n] == 0x3f3f3f3f) return -1;  为什么不这样写呢 存在负权边，前一点到不了，后一点也自然到不了，
    //但是，每个点初始化时0x3f3f3f3f，后一点后被更新  0x3f3f3f3f-x ,这样用==就判断不出来了
    return dist[n];
}

int main()
{
    scanf("%d%d%d", &n, &m, &k);

    for (int i = 0; i < m; i++)
    {
        int a, b, w;
        scanf("%d%d%d", &a, &b, &w);
        edges[i] = {a, b, w};
    }

    int t = bellman_ford();

    if (t == -1) puts("impossible");
    else printf("%d\n", t);

    return 0;
}

作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/123582/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

SPFA

AcWing 851. spfa求最短路

//大部分正权图，就SPFA也可以过掉
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>

using namespace std;

typedef pair<int, int> PII;

const int N = 100010;

int n, m;
int h[N], w[N], e[N], ne[N], idx;
int dist[N];
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

int spfa()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;

    queue<int> q;
    q.push(1);
    st[1] = true;  //st[]存储当前这个点，是否在队列中

    while (q.size())
    {
        int t = q.front();
        q.pop();

        st[t] = false;

        for (int i = h[t]; i != -1; i = ne[i])
        {
            int j = e[i];
            if (dist[j] > dist[t] + w[i])
            {
                dist[j] = dist[t] + w[i];
                if (!st[j])
                {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }

    if (dist[n] == 0x3f3f3f3f) return -1;
    return dist[n];

}

int main()
{
    scanf("%d%d", &n, &m);

    memset(h, -1, sizeof h);

    while (m--)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c);
    }

    int t = spfa();

    if (t == -1) puts("impossible");
    else printf("%d\n", t);

    return 0;
}


作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/123591/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

AcWing 852. spfa判断负环

#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 100010;

int n, m;
int h[N], w[N], e[N], ne[N], idx;
int dist[N], cnt[N];
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

int spfa()
{
    queue<int> q;

    for (int i = 1; i<= n; i++)
    {
        st[i] = true;
        q.push(i);
    }

    while (q.size())
    {
        int t = q.front();
        q.pop();

        st[t] = false;

        for (int i = h[t]; i != -1; i = ne[i])
        {
            int j = e[i];
            if (dist[j] > dist[t] + w[i])
            {
                dist[j] = dist[t] + w[i];
                cnt[j] = cnt[t] + 1;

                if (cnt[j] >= n) return true;
                if (!st[j])
                {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }

    return false;
}

int main()
{
    scanf("%d%d", &n, &m);

    memset(h, -1, sizeof h);

    while (m--)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c);
    }

    if (spfa()) puts("Yes");
    else puts("No");

    return 0;
}

作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/123616/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

Floyd

AcWing 854. Floyd求最短路

#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 210, INF = 1e9;

int n, m, Q;
int d[N][N];

void floyd()
{
    for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}

int main()
{
    scanf("%d%d%d", &n, &m, &Q);

    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            if (i == j) d[i][j] = 0;
            else d[i][j] = INF;

    while (m--)
    {
        int a, b, w;
        scanf("%d%d%d", &a, &b, &w);

        d[a][b] = min(d[a][b], w);
    }

    floyd();

    while (Q--)
    {
        int a, b;
        scanf("%d%d", &a, &b);

        if (d[a][b] > INF / 2) puts("impossible");
        else printf("%d\n", d[a][b]);
    }

    return 0;
}

作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/123753/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

Prim

AcWing 858. Prim算法求最小生成树

#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 510, INF = 0x3f3f3f3f;

int n, m;
int g[N][N];
int dist[N];
bool st[N];

int prim()
{
    memset(dist, 0x3f, sizeof dist);

    int res = 0;
    for (int i = 0; i < n; i++)
    {
        int t = -1;
        for (int j = 1; j <= n; j++)
            if (!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;

        if (i && dist[t] == INF) return INF;
        if (i) res += dist[t];

        for (int j = 1; j <= n; j++) dist[j] = min(dist[j], g[t][j]);
        //if (i) res += dist[t];
        //res不能在这个位置，对于坑点数据 4 4 -10
        //4这个点存在负环，当t==j的时候，min(dist[t], g[t][t])会把自己更新小了
        st[t] = true;  //下标容易打错
    }
    return res;
}

int main()
{
    scanf("%d%d", &n, &m);

    memset(g, 0x3f, sizeof g);

    while (m--)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        g[a][b] = g[b][a] = min(g[a][b], c);  //无向图是一种特殊的有向图
    }

    int t = prim();

    if (t == INF) puts("impossible");
    else printf("%d\n", t);

    return 0;
}

作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/123844/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

Kruskal

AcWing 859. Kruskal算法求最小生成树

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 200010;//SE问题，也可能出现在这里，一般就是边界问题

int n, m;
int p[N];

struct Edge
{
    int a, b, w;

    bool operator< (const Edge &W)const
    {
        return w < W.w;
    }
}edges[N];

int find(int x)
{
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main()
{   
    scanf("%d%d", &n, &m);

    for (int i = 0; i < m; i++)
    {
        int a, b, w;
        scanf("%d%d%d", &a, &b, &w);
        edges[i] = {a, b, w};
    }

    //直接写kruscal, 不用把他单独包起来了
    sort(edges, edges + m);

    for (int i = 1; i <= n; i++) p[i] = i;

    int res = 0, cnt = 0;
    for (int i = 0; i < m; i++)
    {
        int a = edges[i].a, b = edges[i].b, w = edges[i].w;

        a = find(a), b = find(b);
        if (a != b)
        {
            p[a] = b;
            res += w;//最小生成树，所有树边的权重之和
            cnt++; //当前加了多少条边
        }
    }

    if (cnt < n - 1) puts("impossible");
    else printf("%d\n", res);

    return 0;
}

作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/123869/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

染色法判定二分图

AcWing 860. 染色法判定二分图

#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 100010, M = 200010;//无向图，边要多一倍

int n, m;
int h[N], e[M], ne[M], idx;
int color[N];

void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

bool dfs(int u, int c)
{
    color[u] = c;

    for (int i = h[u]; i != -1; i = ne[i])
    {
        int j = e[i];
        if (!color[j])
        {
            if (!dfs(j, 3 - c)) return false;  //把1变成2，把2变成1，就用3-c
        }
        else if (color[j] == c) return false;
    }

    return true;   
}

int main()
{
    scanf("%d%d", &n, &m);

    memset(h, -1, sizeof h);

    while (m--)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b), add(b, a);
    }

    bool flag = true;
    for (int i = 1; i <= n; i++)
        if (!color[i])
        {
            if (!dfs(i ,1))  //1号颜色
            {
                flag = false;
                break;
            }
        }

    if (flag) puts("Yes");
    else puts("No");

    return 0;
}

作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/123879/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

匈牙利算法

AcWing 861. 二分图的最大匹配

#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 510, M = 100010;

int n1, n2, m;
//int h[N], e[N], ne[N], idx;//数组越界，什么错误都可能发生
int h[N], e[M], ne[M], idx;
int match[N];
bool st[N];

void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

bool find(int x)
{
    for (int i = h[x]; i != -1; i = ne[i])
    {
        int j = e[i];
        if (!st[j])
        {
            st[j] = true;
            if (match[j] == 0 || find(match[j]))
            {
                match[j] = x;
                return true;
            }
        }
    }

    return false;
}

int main()
{
    scanf("%d%d%d", &n1, &n2, &m);

    memset(h, -1, sizeof h);

    while (m--)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b);//虽然是无向图，但是匹配找的时候，只会单向的去找
    }

    int res = 0;
    for (int i = 1; i <= n1; i++)
    {
        memset(st, false, sizeof st);
        if (find(i)) res++;
    }

    printf("%d\n", res);

    return 0;
}

作者：zqiceberg
链接：https://www.acwing.com/activity/content/code/content/123892/
来源：AcWing
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单源最短路的建图方式

单源最短路的综合应用

单源最短路的扩展应用

Floyd算法**

最小生成树

最小生成树的扩展应用

负环

差分约束

最近公共祖先

有向图的强连通分量

无向图的双连通分量

二分图

欧拉回路和欧拉路径

拓扑排序**

3 评论

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zzw1   2020-01-17 10:10         踩      回复

还有这种操作？hh

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zqiceberg   2020-01-17 22:25         踩      回复

真的是自己的笔记记录用的啊，没想着怎么分享点赞啊，hh

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zzw1   2020-01-17 22:29    回复了 zqiceberg 的评论         踩      回复

可以的，加油