BZOJ 1123: [POI2008]BLO

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

 #include <cmath>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 #include <algorithm>

 using namespace std;

 #define N 5000000
 #define LL long long

 int n, m; LL ans[N];

 int hd[N], to[N], nt[N], tot;

 int dfn[N], low[N], tim;

 int tarjan(int u, int f)
 {
     dfn[u] = low[u] = ++tim; ans[u] = (n - ) << ;

     int cnt = , siz; LL sum = , tmp = ;

     for (int i = hd[u]; ~i; i = nt[i])if (f != to[i])
     {
         if (!dfn[to[i]])
         {
             siz = tarjan(to[i], u);
             low[u] = min(low[u], low[to[i]]);
             if (low[to[i]] >= dfn[u])
                 ans[u] += 2LL * siz * sum, sum += siz;
             else
                 tmp += siz;
         }
         else
             low[u] = min(low[u], dfn[to[i]]);
     }

     ans[u] += 2LL * (n - (sum + )) * sum;

     return sum + tmp + ;
 }

 signed main(void)
 {
     scanf("%d%d", &n, &m);

     memset(hd, -, sizeof(hd)), tot = ;

     for (int i = ; i <= m; ++i)
     {
         int x, y; scanf("%d%d", &x, &y);

         nt[tot] = hd[x]; to[tot] = y; hd[x] = tot++;
         nt[tot] = hd[y]; to[tot] = x; hd[y] = tot++;
     }

     memset(dfn, , sizeof(dfn)); tim = ; tarjan(, -);

     for (int i = ; i <= n; ++i)
         printf("%lld\n", ans[i]);
 }

BZOJ_1123.cpp

 #include <cstdio>

 template <class T>
 inline T min(const T &a, const T &b)
 {
     return a < b ? a : b;
 }

 typedef long long lnt;

 const int mxn = ;
 const int mxm = ;

 int n, m;

 int hd[mxn];
 int to[mxm];
 int nt[mxm];

 int dfn[mxn];
 int low[mxn];

 lnt ans[mxn];

 lnt tarjan(int u, int f)
 {
     static int tim = ;

     ans[u] = (n - ) << ;
     dfn[u] = low[u] = ++tim;

     lnt siz, sum = , tmp = ;

     for (int i = hd[u], v; i; i = nt[i])
         if ((v = to[i]) != f)
         {
             if (!dfn[v])
             {
                 siz = tarjan(v, u);

                 low[u] = min(low[u], low[v]);

                 if (low[v] >= dfn[u])
                     ans[u] += 2LL * sum * siz, sum += siz;
                 else
                     tmp += siz;
             }
             else
                 low[u] = min(low[u], dfn[v]);
         }

     ans[u] += 2LL * (n - sum - ) * sum;

     return sum + tmp + ;
 }

 signed main(void)
 {
     scanf("%d%d", &n, &m);

     for (int i = ; i < m; ++i)
     {
         static int x, y, tot;

         scanf("%d%d", &x, &y);

         nt[++tot] = hd[x], to[tot] = y, hd[x] = tot;
         nt[++tot] = hd[y], to[tot] = x, hd[y] = tot;
     }

     tarjan(, );

     for (int i = ; i <= n; ++i)
         printf("%lld\n", ans[i]);
 }

@Author: YouSiki