UVA 10765 Doves and bombs

给定一个无向的连通图，要求每个点去掉后连通分量的数目，然后输出连通分量最多的m个点。

分析：
先求出双连通分量，然后统计所有双连通分量中割顶出现的次数，最后求出的就是割顶去掉后剩下的双连通的数目，对于不是割顶的点，去掉后剩下的仍为双连通，所以结果就是1.

代码：

#include <iostream>
#include <sstream>
#include <cstdio>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <string>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#define esp 1e-6
#define pi acos(-1.0)
#define pb push_back
#define mp(a, b) make_pair((a), (b))
#define in  freopen("in.txt", "r", stdin);
#define out freopen("out.txt", "w", stdout);
#define print(a) printf("%d\n",(a));
#define bug puts("********))))))");
#define stop  system("pause");
#define Rep(i, c) for(__typeof(c.end()) i = c.begin(); i != c.end(); i++)

#define inf 0x0f0f0f0f

using namespace std;
typedef long long  LL;
typedef vector<int> VI;
typedef pair<int, int> pii;
typedef vector<pii,int> VII;
typedef vector<int>:: iterator IT;
const int maxn = 10000 + 10;
int pre[maxn], iscut[maxn], low[maxn], bccno[maxn], dfs_clock, bcc_cnt;
VI g[maxn], bcc[maxn];
int ans[maxn], r[maxn];
stack<pii> S;
bool cmp (const int &a, const int &b)
{
    return ans[a] > ans[b] || (ans[b] == ans[a] && a < b);
}
int dfs(int u, int fa)
{
    int lowu = pre[u] = ++dfs_clock;
    int child = 0;
    for(int i = 0; i < g[u].size(); i++)
    {
        int v = g[u][i];
        if(!pre[v])
        {
            S.push(mp(u, v));
            child++;
            int lowv = dfs(v, u);
            lowu = min(lowu, lowv);
            if(lowv >= pre[u])
            {
                iscut[u] = 1;
                bcc_cnt++, bcc[bcc_cnt].clear();
                for(;;)
                {
                    pii x = S.top();
                    S.pop();
                    if(bccno[x.first] != bcc_cnt)
                    {
                        bccno[x.first] = bcc_cnt;
                        bcc[bcc_cnt].pb(x.first);
                    }
                    if(bccno[x.second] != bcc_cnt)
                    {
                        bccno[x.second] = bcc_cnt;
                        bcc[bcc_cnt].pb(x.second);
                    }
                    if(x.first == u && x.second == v) break;
                }
            }
        }
        else if(pre[v] < pre[u] && v != fa)
        {
            S.push(mp(u, v));
            lowu = min(lowu, pre[v]);
        }
    }
    if(child == 1 && fa < 0) iscut[u] = 0;
    return low[u] = lowu;
}
void find_bcc(int n)
{
    memset(pre, 0, sizeof(pre));
    memset(iscut, 0, sizeof(iscut));
    memset(bccno, 0, sizeof(bccno));
    dfs_clock = bcc_cnt = 0;

    for(int i = 0; i < n; i++)
        if(!pre[i])
            dfs(i, -1);
}
int main(void)
{

    int n, m;
    while(scanf("%d%d", &n, &m), n||m)
    {
        for(int i = 0; i < maxn; i++)
            g[i].clear();
        for(int i = 0; i < maxn; i++)
            r[i] = i;
        memset(ans, 0,sizeof(ans));
        int u, v;
        while(scanf("%d%d", &u, &v), u>=0)
        {
            g[u].pb(v);
            g[v].pb(u);
        }
        find_bcc(n);
        for(int i = 1; i <= bcc_cnt; i++)
            for(int j = 0; j < bcc[i].size(); j++)
            {
                int x = bcc[i][j];
                if(iscut[x])
                    ans[x]++ ;
            }
        for(int i = 0; i < n; i++)
            if(ans[i] == 0)
                ans[i] = 1;
        sort(r, r+n, cmp);
        for(int i = 0; i < m; i++)
            printf("%d %d\n", r[i], ans[r[i]]);
        puts("");
    }
    return 0;
}