BZOJ 1123: [POI2008]BLO( tarjan )

tarjan找割点..不是割点答案就是(N-1)*2, 是割点的话就在tarjan的时候顺便统计一下

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#include<cstdio>

#include<cstring>

#include<algorithm>

 

using namespace std;

 

typedef long long ll;

 

const int maxn = 100009;

const int maxm = 500009;

 

struct edge {

int to;

edge* next;

} E[maxm << 1], *pt = E, *head[maxn];

 

void add(int u, int v) {

pt->to = v; pt->next = head[u]; head[u] = pt++;

}

void addedge(int u, int v) {

add(u, v); add(v, u);

}

 

edge* Rev(edge* e) {

return E + ((e - E) ^ 1);

}

 

int dfn[maxn], low[maxn], size[maxn], CK = 0, N;

ll ans[maxn];

 

void tarjan(int x, edge* r) {

dfn[x] = low[x] = ++CK;

size[x] = 1;

int cnt = 0;

for(edge* e = head[x]; e; e = e->next) if(e != r) {

if(!dfn[e->to]) {

tarjan(e->to, Rev(e));

if(low[e->to] >= dfn[x])

ans[x] += cnt * size[e->to], cnt += size[e->to];

else

low[x] = min(low[x], low[e->to]);

size[x] += size[e->to];

} else

low[x] = min(low[x], dfn[e->to]);

}

ans[x] += ll(cnt) * (N - cnt - 1);

}

 

void init() {

int m;

scanf("%d%d", &N, &m);

while(m--) {

int u, v; scanf("%d%d", &u, &v);

addedge(--u, --v);

}

}

 

void work() {

memset(ans, 0, sizeof ans);

memset(dfn, 0, sizeof dfn);

memset(low, 0, sizeof low);

for(int i = 0; i < N; i++)

if(!dfn[i]) tarjan(i, NULL);

for(int i = 0; i < N; i++)

printf("%lld\n", 2LL * (ans[i] + N - 1));

}

 

int main() {

init();

work();

return 0;

}

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1123: [POI2008]BLO

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 751  Solved: 309
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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