UVaLive 7456 Least Crucial Node (并查集+暴力 或者 求割点)

题意：求标号最小的最大割点.(删除该点后，指定点#sink能到达的点数减少最多).

析：由于不知道要去掉哪个结点，又因为只有100个结点，所以我们考虑用一个暴力，把所有的结点都去一次，然后用并查集去判断。

当然也可以用割点和桥的模板，最后再判断一下，哪个点后面的点有多少就好。

代码如下：

并查集+暴力：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 100;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int x;
vector<P> v;
int p[105];
int Find(int x) {  return x == p[x] ? x : p[x] = Find(p[x]); }

int main(){
    while(scanf("%d", &n) == 1 && n){
        scanf("%d", &x);
        scanf("%d", &m);
        int u, vv;
        v.clear();
        for(int i = 0; i < m; ++i){
            scanf("%d %d", &u, &vv);
            v.push_back(P(u, vv));
        }
        int ans = 0, cnt = 0;
        for(int i = 1; i <= n; ++i){
            if(i == x)  continue;
            for(int j = 1; j <= n; ++j)  p[j] = j;
            for(int j = 0; j < v.size(); ++j){
                u = v[j].first;
                vv = v[j].second;
                if(u == i || vv == i)  continue;
                int x = Find(u);
                int y = Find(vv);
                if(x != y)  p[y] = x;
            }
            map<int, int> mp;
            map<int, int> :: iterator it;
            for(int j = 1; j <= n; ++j)
                if(i != j)  ++mp[Find(j)];
            if(mp.size() <= 1) continue;
            int y = Find(x);
            if(cnt < n-mp[y]-1){
                cnt = n-mp[y]-1;
                ans = i;
            }
        }

        printf("%d\n", ans);
    }
    return 0;
}

割点和桥：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e2 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
vector<int> G[maxn];
bool cut[maxn];
int low[maxn], dfn[maxn], vis[maxn], ans[maxn];

void dfs(int cur, int fa, int d){
    vis[cur] = 1;
    dfn[cur] = low[cur] = d;
    int u = 0;
    for(int i = 0; i < G[cur].size(); ++i){
        int v = G[cur][i];
        if(v != fa && 1 == vis[v]){
            if(dfn[v] < low[cur])  low[cur] = dfn[v];
        }
        if(!vis[v]){
            dfs(v, cur, d+1);
            ++u;
            if(low[v] < low[cur])  low[cur] = low[v];
            if((fa == -1 && u > 1) || (fa != -1 && low[v] >= dfn[cur]))  cut[cur] = true;
        }
    }
    vis[cur] = 2;
}

void dfs1(int u){
    vis[u] = 1;
    ans[u] = 1;
    for(int i = 0; i < G[u].size(); ++i){
        int v = G[u][i];
        if(vis[v])  continue;
        dfs1(v);
        ans[u] += ans[v];
    }
}

int main(){
    int rt;
    while(scanf("%d", &n) == 1 && n){
        scanf("%d", &rt);
        scanf("%d", &m);
        for(int i = 1; i <= n; ++i)  G[i].clear();
        int u, v;
        for(int i = 0; i < m; ++i){
            scanf("%d %d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        memset(dfn, 0, sizeof dfn);
        memset(cut, false, sizeof cut);
        memset(low, 0, sizeof vis);
        memset(vis, 0, sizeof vis);
        dfs(rt, -1, 0);
        memset(vis, 0, sizeof vis);
        memset(ans, 0, sizeof ans);
        dfs1(rt);

        int anss = 0, cnt = 0;
        for(int i = 1; i <= n; ++i){
            if(cut[i] && cnt < ans[i]){
                cnt = ans[i];
                anss = i;
            }
        }
        printf("%d\n", anss);
    }
    return 0;
}