Description
给你一张无向连通图，每次有一个操作，将x~y这条路径上的必经点的权值全部加一。
最后输出每个点权值。（所有权值初始为0）

Sample Input
4 4 2
1 2
1 3
2 3
1 4
4 2
4 3

Sample Output
2
1
1
2

做了这道题感觉又回到了煜东神的怀抱
那你不是缩一下点，树上差分一下就完事了吗。。。

#include <vector>
#include <cstdio>
#include <cstring>

using namespace std;
int _min(int x, int y) {return x < y ? x : y;}
int read() {
    int s = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
    return s * f;
}

struct edge {
    int x, y, next;
} e[410000], e2[410000]; int len, len2, last[210000], last2[210000];
int id, cnt, low[210000], dfn[210000];
int tp, sta[210000]; bool cut[210000];
int belong[210000], ans[210000], s[210000];
int dep[210000], fa[210000][20];
vector<int> q[210000];

void ins(int x, int y) {
    e[++len].x = x; e[len].y = y;
    e[len].next = last[x]; last[x] = len;
}

void ins2(int x, int y) {
    e2[++len2].x = x; e2[len2].y = y;
    e2[len2].next = last2[x]; last2[x] = len2;
}

void tarjan(int x) {
    dfn[x] = low[x] = ++id;
    sta[++tp] = x; int uu = 0;
    for(int k = last[x]; k; k = e[k].next) {
        int y = e[k].y;
        if(!dfn[y]) {
            tarjan(y);
            low[x] = _min(low[x], low[y]);
            if(low[y] >= dfn[x]) {
                uu++;
                if(x != 1 || uu > 1) cut[x] = 1;
                int i; cnt++;
                do {
                    i = sta[tp--];
                    q[cnt].push_back(i);
                } while(i != y);
                q[cnt].push_back(x);
            }
        } else low[x] = _min(low[x], dfn[y]);
    }
}

void dfs(int x) {
    for(int i = 1; (1 << i) <= dep[x]; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
    for(int k = last2[x]; k; k = e2[k].next) {
        int y = e2[k].y;
        if(y != fa[x][0]) {
            fa[y][0] = x;
            dep[y] = dep[x] + 1;
            dfs(y);
        }
    }
}

int LCA(int x, int y) {
    if(dep[x] > dep[y]) swap(x, y);
    for(int i = 18; i >= 0; i--) if(dep[y] - dep[x] >= (1 << i)){
        y = fa[y][i];
    } if(x == y) return x;
    for(int i = 18; i >= 0; i--) if(fa[x][i] != fa[y][i]){
        x = fa[x][i], y = fa[y][i];
    } return fa[x][0];
}

void gets(int x) {
    for(int k = last2[x]; k; k = e2[k].next) {
        int y = e2[k].y;
        if(fa[x][0] != y) {
            gets(y); s[x] += s[y];
        }
    }
}

int main() {
    int n = read(), m = read(), Q = read();
    for(int i = 1; i <= m; i++) {
        int x = read(), y = read();
        ins(x, y), ins(y, x);
    }
    tarjan(1);
    int num = cnt;
    for(int i = 1; i <= n; i++) if(cut[i]) belong[i] = ++num;
    for(int i = 1; i <= cnt; i++) {
        for(int j = 0; j < q[i].size(); j++) {
            int x = q[i][j];
            if(cut[x]) {
                ins2(belong[x], i);
                ins2(i, belong[x]);
            } else belong[x] = i;
        }
    } dfs(1);
    for(int i = 1; i <= Q; i++) {
        int x = read(), y = read(); ans[x]++; ans[y]++;
        x = belong[x], y = belong[y];
        int lca = LCA(x, y);
        s[x]++; s[y]++;
        s[lca]--, s[fa[lca][0]]--;
    } gets(1);
    for(int i = 1; i <= n; i++) {
        if(cut[i]) printf("%d\n", s[belong[i]]);
        else printf("%d\n", ans[i]);
    }
    return 0;
}