bzoj1123 [POI2008]BLO

[POI2008]BLO

Time Limit: 10 Sec Memory Limit: 162 MB

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

<br >
<br >
<br >
<br >
<br >

这可能是个割点吧。。。。网上的博客写的都特别好。。。
大概就是简单计数

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
struct lpl{
    int fa, size, low, dfn;
}node[maxn];
vector<int> point[maxn];
int n, m, cnt, root = 1, llppdd;
long long ans[maxn];
bool flag[maxn];

inline void putit()
{
    int u, v;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; ++i){
        scanf("%d%d", &u, &v);
        point[u].push_back(v); point[v].push_back(u);
    }
}

void dfs(int t)
{
    node[t].dfn = node[t].low = ++cnt;
    node[t].size = 1; llppdd = point[t].size() - 1;
    long long lin = 0;
    for(int i = point[t].size() - 1; i >= 0; --i){
        int now = point[t][i];
		if(node[now].dfn) node[t].low = min(node[t].low, node[now].dfn);
        else{
            node[now].fa = t; dfs(now);
            node[t].size += node[now].size;	node[t].low = min(node[t].low, node[now].low);	
			if(node[t].dfn <= node[now].low){
				ans[t] += (long long)lin * node[now].size;
				lin += node[now].size;
			}
        }
    }        
	ans[t] += (long long)lin * (n - 1 - lin);
}

int main()
{
    putit();
    dfs(root);
    for(int i = 1; i <= n; ++i) printf("%lld
", (ans[i] + n - 1) * 2);
    return 0;
}