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

using namespace std;
const int mod = 47;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    string a, b;
    cin >> a >> b;

    long long x = 1, y = 1;
    for (int i = 0;i < a.size();i ++) x = x * (a[i] - 'A' + 1) % mod;
    for (int i = 0;i < b.size();i ++) y = y * (b[i] - 'A' + 1) % mod;

    if (x == y) puts("GO");
    else puts("STAY");
    return 0;
}

一开始都设置成0x3f3f3f3f可以一条一条拓展出去，让每次贪心走在有效的路线

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

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

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



    for (int i = 1; i <= n; i ++ )
    {
        int t = -1;//这个记得写进循环里边，因为每次都要贪心最近的那个编号

        for (int j = 1; j <= n; j ++ )
            if (! st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;
        if (t == n) break;
        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()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    memset(g, 0x3f,sizeof g);
    cin >> n >> m;

    while (m -- )
    {
        int a, b, d;
        cin >> a >> b >> d;
        g[a][b] = min(g[a][b], d);
    }

    cout << dijkstra() << "\n";
    return 0;
}

感觉听课学到的不多，顶多只是给个方向

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

using namespace std;
const int N = 1e5 + 10;

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

int 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] == 0) q[++ tt] = i;

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

    return tt == n - 1;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);

    memset(h, -1, sizeof h);

    cin >> n >> m;
    while (m -- )
    {
        int a, b;
        cin >> a >> b;
        add(a, b);
        d[b] ++;
    }

    if(topsort())
    {
        for (int i = 0; i < n; i ++ )
            cout << q[i] << ' ';
        puts("");
    }
    else puts("-1");

    return 0;
}

cy

cy

看了半天后面忍不住看题解才发现公式没错，当时差分形式写错了。注意正确的差分是

b[l] += c;
b[r + 1] -= c; //这里记得＋1，要等下标跑出去的时候才减掉它

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

using namespace std;
const int N = 2e5 + 10;

int a[N];
int n, m, k;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);

    cin >> n >> m >> k;
    while (n -- )
    {
        int ti, ci;
        cin >> ti >> ci;
        int l = ti - k - ci + 1,r = ti - k;
        if(r > 0)
        {
            a[max(1,l)] ++;
            a[r + 1] --;
        }
    }

    for (int i = 1; i <= N; i ++ ) a[i] += a[i - 1];

    while (m -- )
    {
        int t;
        cin >> t;
        cout << a[t] << "\n";
    }

    return 0;
}

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

using namespace std;

const int N = 33, M = 300010;

int n, x;
int w[N], f[M];

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

    int sum = 0;
    for (int i = 0; i < n; i ++ )
    {
        scanf("%d", &w[i]);
        sum += w[i];
    }

    int m = sum - x;
    for (int i = 0; i < n; i ++ )
        for (int j = m; j >= w[i]; j -- )
            f[j] = max(f[j], f[j - w[i]] + w[i]);

    printf("%d\n", sum - f[m]);
    return 0;
}

dfs要处理好末尾情况和普通情况的协调，要提起归纳好
bfs要处理好特殊情况的返回

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10;
int h[N],e[N],ne[N],idx;
int d[N],q[N];
int n, m;

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()
{
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);

    memset(h, -1, sizeof h);

    cin >> n >> m;
    while (m -- )
    {
        int a, b;
        cin >> a >> b;
        add(a, b);
    }

    cout << bfs() << "\n";
    return 0;
}

#include<bits/stdc++.h>

using namespace std;
const int N = 1e5 + 10;

int h[N],ne[2 * N],e[2 * N];
int idx;
bool st[N];
int n;
int ans = N;

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

int dfs(int u)
{
    int res = 0;
    st[u] = true; //标记这个点已经成为父亲，所以遍历的时候不会穿过这个节点

    int sum = 1; //这个sum是给后面求头顶上面那一堆连通图准备的

    for (int i = h[u];i != -1;i = ne[i]) //h[i]存的是门牌
    {
        int j = e[i]; //e[i]存的才是点本身
        if(!st[j])
        {
            int s = dfs(j);
            //到这里下面的那些数据已经处理好了，s返回的是子树的结点数
            res = max(res , s); //去掉没用的数据，只取子树连接点数最大的那个
            sum += s; //这个sum先加起来，依然是给头顶那一堆准备的
        }

    }

    res = max(res, n - sum);
    ans = min(ans, res); //每次都更新一下答案

    return sum; //这里返回的是子树的大小，这样上面的父亲就可以继续sum += s;来运作这个递归
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);

    memset(h, -1, sizeof h);

    cin >> n; //有n个节点，树没有环，可以移动任意两个最后能归于一条线，那就是n - 1条边
    for (int i = 0; i < n - 1; i ++ )
    {
        int a, b;
        cin >> a >> b;
        add(a, b);
        add(b, a);
    }

    dfs(1);
    cout << ans << "\n";
    return 0;
}