[BZOJ]1123 [POI2008]BLO 割点

1123: [POI2008]BLO

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 1301  Solved: 591
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Description

Byteotia城市有n个 towns m条双向roads. 每条 road 连接 两个不同的 towns ,没有重复的road. 所有towns连通。

Input

输入n<=100000 m<=500000及m条边

Output

输出n个数，代表如果把第i个点去掉，将有多少对点不能互通。

Sample Input

5 5
1 2
2 3
1 3
3 4
4 5

Sample Output

8
8
16
14
8

HINT

Source

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    割点板子题.

    在tarjan的时候记录一下size即可. 注意乘2.

#include<stdio.h>#include<algorithm>using namespace std;typedef long long dnt;const int maxn = 100005;dnt ans[maxn];int n, m, num, indexx;int low[maxn], dfn[maxn], siz[maxn], h[maxn];struct edge{ int nxt, v;}e[1000005];inline void add(int u, int v){e[++num].v = v, e[num].nxt = h[u], h[u] = num;e[++num].v = u, e[num].nxt = h[v], h[v] = num;}inline void dfs(int u){siz[u] = 1;dfn[u] = low[u] = ++indexx;int sum = 0;for(int i = h[u]; i; i = e[i].nxt){int v = e[i].v;if(dfn[v]) low[u] = min(low[u], dfn[v]);else{dfs(e[i].v);siz[u] += siz[v];low[u] = min(low[u], low[v]);if(low[v] >= dfn[u]){ans[u] += (dnt) sum * siz[v];sum += siz[v];}}}ans[u] += (dnt) sum * (n - sum - 1);}int main(){scanf("%d%d", &n, &m);for(int i = 1; i <= m; ++i){int u, v;scanf("%d%d", &u, &v);add(u, v);}dfs(1);for(int i = 1; i <= n; ++i) printf("%lld\n", (ans[i] + n - 1) * 2);}