1123: [POI2008]BLO

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 1030  Solved: 440
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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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分析

如果一个点不是割点，那么删去后不会对其他点之间的连通性造成影响；如果是一个割点，影响只和其连接的几个块的大小有关。因此Tarjan求割点的同时注意维护子树大小即可。

代码

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <iostream>
 6 #include <algorithm>
 7
 8 using namespace std;
 9
10 #define N 5000000
11 #define LL long long
12
13 int n, m; LL ans[N];
14
15 int hd[N], to[N], nt[N], tot;
16
17 int dfn[N], low[N], tim;
18
19 int tarjan(int u, int f)
20 {
21     dfn[u] = low[u] = ++tim; ans[u] = (n - 1) << 1;
22
23     int cnt = 0, siz; LL sum = 0, tmp = 0;
24
25     for (int i = hd[u]; ~i; i = nt[i])if (f != to[i])
26     {
27         if (!dfn[to[i]])
28         {
29             siz = tarjan(to[i], u);
30             low[u] = min(low[u], low[to[i]]);
31             if (low[to[i]] >= dfn[u])
32                 ans[u] += 2LL * siz * sum, sum += siz;
33             else
34                 tmp += siz;
35         }
36         else
37             low[u] = min(low[u], dfn[to[i]]);
38     }
39
40     ans[u] += 2LL * (n - (sum + 1)) * sum;
41
42     return sum + tmp + 1;
43 }
44
45 signed main(void)
46 {
47     scanf("%d%d", &n, &m);
48
49     memset(hd, -1, sizeof(hd)), tot = 0;
50
51     for (int i = 1; i <= m; ++i)
52     {
53         int x, y; scanf("%d%d", &x, &y);
54
55         nt[tot] = hd[x]; to[tot] = y; hd[x] = tot++;
56         nt[tot] = hd[y]; to[tot] = x; hd[y] = tot++;
57     }
58
59     memset(dfn, 0, sizeof(dfn)); tim = 0; tarjan(1, -1);
60
61     for (int i = 1; i <= n; ++i)
62         printf("%lld\n", ans[i]);
63 }

BZOJ_1123.cpp

@Author: YouSiki