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;

}

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

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巴特西

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

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