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#include <bits/stdc++.h>

using namespace std;

const int N = 100010, M = 200020;
int n, m, p[N];

struct E
{
    int a, b, c;
    bool operator < (const E& w)
    {
        return c < w.c;
    }
}e[M];

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


int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i ++) p[i] = i;
    int res = 0, cnt = 0;
    for(int i = 0; i < m; i ++)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        e[i] = {a, b, c};
    }
    sort(e, e + m);
    for(int i = 0; i < m; i ++)
    {
        int a = e[i].a, b = e[i].b, w = e[i].c;
        if(find(a) != find(b)) p[p[a]] = p[b], cnt ++, res += w;
    }
    if(cnt != n - 1) puts("impossible");
    else printf("%d\n", res);
    return 0;
}



#include <iostream>
#include <cstring>

using namespace std;

const int N = 510, INF = 0x3f3f3f3f;
int g[N][N], dist[N],n, m;
bool st[N];

int prim()
{
    int res = 0;
    memset(dist, 0x3f, sizeof dist);
    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;
        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]);
    }
    return res;
}

int main()
{
    memset(g, 0x3f, sizeof g);
    cin >> n >> m;
    while(m --)
    {
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = g[b][a] = min(g[a][b], c);
    }
    int t = prim();
    if(t == INF) puts("impossible");
    else printf("%d\n", t);
    return 0;
}



活动打卡代码 AcWing 854. Floyd求最短路

floyd算法,三重循环,内层一定是d[i][j] 和 d[i][k] + d[k][j],的最小值,因为是i到j的距离。

#include <iostream>
#include <cstring>

using namespace std;

const int N = 210;
int d[N][N], n, m, k;

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()
{
    cin >> n >> m >> k;
    memset(d, 0x3f, sizeof d);
    for(int i = 1; i <= n; i ++) d[i][i] = 0;
    while(m --)
    {
        int a, b, c;
        cin >> a >> b >> c;
        d[a][b] = min(d[a][b], c);
    }
    floyd();
    while(k --)
    {
        int a, b;
        cin >> a >> b;
        if(d[a][b] > 0x3f3f3f3f / 2) puts("impossible");
        else cout << d[a][b] << endl;
    }
    return 0;
}


活动打卡代码 AcWing 852. spfa判断负环

不需要初始化距离是因为存在负环一定会在负环内更新,虽然dist数组内容不正确,但是不影响判断负环的正确性。

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

using namespace std;

const int N = 100010;
int h[N], ne[N], e[N], idx;
int n, m, dist[N], cnt[N], w[N];
bool st[N];

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

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

    while(q.size())
    {
        auto 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])
            {
                cnt[j] = cnt[t] + 1;
                dist[j] = dist[t] + w[i];
                if(cnt[j] >= n) return true;
                if(!st[j])
                {
                    st[j] = true;
                    q.push(j);
                }
            }
        }
    }
    return false;
}

int main()
{
    memset(h, -1, sizeof h);
    cin >> n >> m;
    while(m --)
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }
    if(spfa()) puts("Yes");
    else puts("No");
    return 0;
}


活动打卡代码 AcWing 851. spfa求最短路

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

using namespace std;

const int N = 100010, M = 200010;
int h[N], e[M], ne[M], idx, dist[N], w[N], n, m;
bool st[N];

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

int spfa()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    queue<int> q;
    st[1] = true;
    q.push(1);
    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])
                {
                    st[j] = true;
                    q.push(j);
                }
            }
        }
    }
    if(dist[n] == 0x3f3f3f3f) return -1;
    return dist[n];
}

int main()
{
    memset(h, -1, sizeof h);
    cin >> n >> m;
    while(m --)
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }
    int t = spfa();
    if(t == -1) puts("impossible");
    else cout << t << endl;
    return 0;
}



#include <iostream>
#include <cstring>

using namespace std;

const int N = 510, M = 10010;
int dist[N], last[N], n, m, k;
struct edge
{
    int a, b, c;
}edges[M];

void bellman_ford()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    for(int i = 0; i < k; i ++)
    {
        memcpy(last, dist, sizeof dist);
        for(int j = 0; j < m; j ++)
        {
            int a, b, w;
            a = edges[j].a, b = edges[j].b, w = edges[j].c;
            dist[b] = min(dist[b], last[a] + w);
        }
    }
}

int main()
{
    cin >> n >> m >> k;
    for(int i = 0; i < m; i ++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        edges[i] = {a, b, c};
    }
    bellman_ford();
    if(dist[n] > 0x3f3f3f3f / 2)  puts("impossible");
    else cout << dist[n] << endl;
    return 0;
}




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

using namespace std;
typedef pair<int, int> PII;

const int N = 1000010;
int n, m, h[N], ne[N], e[N], idx, w[N], dist[N];
bool st[N];
void add(int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, 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, dis = t.first;
        if(st[ver]) continue;
        st[ver] = true;
        for(int i = h[ver]; i != -1; i = ne[i])
        {
            int j = e[i];
            if(dist[j] > dist[ver] + w[i])
            {
                dist[j] = dis + w[i];
                heap.push({dist[j], j});
            }
        }
    }
    if(dist[n] == 0x3f3f3f3f) return -1;
    return dist[n];
}

int main()
{
    memset(h, -1, sizeof h);
    cin >> n >> m;
    while(m --)
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }
    cout << dijkstra() << endl;
    return 0;
}



题目链接 https://www.acwing.com/problem/content/850/

我想输出所有的拓扑排序。

这是我写的代码,但是不确定对不对,测试了几个貌似还可以。

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

using namespace std;

const int N = 100010;
int q[N], d[N], h[N], e[N], ne[N], idx, n, m;
bool st[N];

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

void dfs(int u)
{
    if(u > n)
    {
        for(int i = 1; i <= n ; i ++) printf("%d ", q[i]);
        puts("");
        return;
    }
    for(int i = 1; i <= n; i ++)
    {
        if(d[i] == 0 && !st[i])
        {
            q[u] = i;
            for(int j = h[i]; j != -1; j = ne[j])
            {
                int t = e[j];
                d[t] --;
            }
            st[i] = true;
            dfs(u + 1);
            st[i] = false;
            for(int j = h[i]; j != -1; j = ne[j])
            {
                int t = e[j];
                d[t] ++;
            }
        }
    }
}

int main()
{
    memset(h, -1, sizeof h);
    cin >> n >> m;
    while(m --)
    {
        int a, b;
        cin >> a >> b;
        add(a, b);
    }
    dfs(1);
    return 0;
}

看一看这个暴搜对不对




#include <iostream>
#include <cstring>

using namespace std;

const int N = 510;
int g[N][N], n, m, dist[N];
bool st[N];

int dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    for(int i = 0; i < n - 1; 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;
    return dist[n];
}

int main()
{
    memset(g, 0x3f, sizeof g);
    cin >> n >> m;
    while(m --)
    {
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = min(g[a][b], c);
    }
    cout << dijkstra() << endl;
    return 0;
}



#include <iostream>
#include <cstring>

using namespace std;

const int N = 100010, M = 200020;
int h[N], e[M], ne[M], q[N], d[N], idx, n, m;

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

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])
        {
            int j = e[i];
            if(-- d[j] == 0) q[++ tt] = j;
        }
    }
    return tt == n - 1;
}

int main()
{
    memset(h, -1, sizeof h);
    cin >> n >> m;
    while(m --)
    {
        int a, b;
        cin >> a >> b;
        add(a, b);
    }
    if(topsort())
    {
        for(int i = 0; i < n; i ++)
        printf("%d ", q[i]);
    }
    else puts("-1");
    return 0;
}