POJ 1523 SPF （无向图求割点）

POJ 1523 SPF

链接：http://poj.org/problem?id=1523

题意：给定一个无向连通图，求割点。并计算出去除每个割点后能将图分为多少块。

思路：裸的求无向图割点。

代码：

/*
ID: [email protected]
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, a, n) for (int i = n - 1; i >= a; i--)
#define eps 1e-6
#define debug puts("===============")
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m)
typedef long long ll;
typedef unsigned long long ULL;
const int maxn = 1100;
vector<int> g[maxn];
int dfn[maxn], low[maxn], vis[maxn], sub[maxn]; //sub记录每个割点分为多少块
bool cut[maxn], bridge[maxn][maxn];
void cut_bridge(int u, int f, int dep) {
    vis[u] = 1; dfn[u] = low[u] = dep;
    int child = 0;
    for (int i = 0; i < g[u].size(); i++) if (g[u][i] != f) {
        int v = g[u][i];
        if (vis[v] == 1 && dfn[v] < low[u]) low[u] = dfn[v];    //回边更新low值
        if (vis[v] == 0) {
            cut_bridge(v, u, dep + 1);
            child++;
            if (low[v] < low[u]) low[u] = low[v];  //通过儿子回到祖先
            if (f == -1 && child > 1) cut[u] = true, sub[u] = child;  //根节点为割点
            if (f != -1 && low[v] >= dfn[u]) cut[u] = true, sub[u]++; //非根节点为割点
            if (low[v] > dfn[u]) bridge[u][v] = bridge[v][u] = true; //割边
        }
    }
    vis[u] = 2;
}
int main () {
    int cas = 1;
    int u, v, mx = 0;
    //freopen("2.txt", "r", stdin);
    while(~scanf("%d", &u), u) {
        mx = max(mx, u);
        for (int i = 0; i < maxn; i++) g[i].clear();
        scanf("%d", &v);
        mx = max(mx, v);
        g[u].push_back(v), g[v].push_back(u);
        while(~scanf("%d", &u), u) {
            mx = max(mx, u);
            scanf("%d", &v);
            mx = max(mx, v);
            g[u].push_back(v), g[v].push_back(u);
        }
        memset(vis, 0, sizeof(vis));
        memset(cut, 0, sizeof(cut));
        memset(bridge, 0, sizeof(bridge));
        memset(sub, 0, sizeof(sub));
        cut_bridge(mx, -1, 0);
        if (cas > 1) puts("");
        printf("Network #%d\n", cas++);
        bool has = false;
        for (int i = 1; i <= mx; i++) if (sub[i]) printf("  SPF node %d leaves %d subnets\n", i, sub[i] + 1), has = true;
        if (!has) puts("  No SPF nodes");
        mx = 0;
    }
    return 0;
}