ACM/ICPC 之 混合图的欧拉回路判定-网络流(POJ1637)

 

//网络流判定混合图欧拉回路
//通过网络流使得各点的出入度相同则possible，否则impossible
//残留网络的权值为可改变方向的次数，即n个双向边则有n次
//Time:157Ms    Memory:348K
#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
#define MAXN 205
#define INF 0x3f3f3f3f
int n,m;
int s,t;
int dif[MAXN];
int res[MAXN][MAXN];    //残留网络-代表可变方向数
int pre[MAXN];
bool bfs()
{
    memset(pre,-1,sizeof(pre));
    queue<int> q;
    q.push(s); pre[s] = 0;
    while(!q.empty()){
        int cur = q.front();
        q.pop();
        for(int i = 1; i <= t; i++)
        {
            if(pre[i] == -1 && res[cur][i])
            {
                pre[i] = cur;
                if(i == t) return true;
                q.push(i);
            }
        }
    }
    return false;
}
int EK()
{
    int maxFlow = 0;
    while(bfs()){
        int mind = INF;
        for(int i = t; i != s; i = pre[i])
            mind = min(mind, res[pre[i]][i]);
        for(int i = t; i != s; i = pre[i])
        {
            res[pre[i]][i] -= mind;
            res[i][pre[i]] += mind;
        }
        maxFlow += mind;
    }
    return maxFlow;
}
int main()
{
    //freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--){
        memset(dif,0,sizeof(dif));
        memset(res,0,sizeof(res));
        scanf("%d%d", &n, &m);
        int total = 0;
        s = 0; t = n+1;
        for(int i = 0; i < m; i++)
        {
            int u,v,t;
            scanf("%d%d%d", &u,&v,&t);
            dif[u]++;   dif[v]--;
            if(t == 0) res[u][v] += 1;  //重边则可变方向+1
        }
        bool flag = true;
        for(int i = 1; i <= n; i++)
        {
            if(dif[i] > 0) {    //出度多-通过源点给予奇数入度
                res[s][i] = dif[i]/2;
                total += dif[i]/2;
            }
            if(dif[i] < 0) res[i][t] = -dif[i]/2;  //入度多-通过汇点给予奇数出度
            if(abs(dif[i]) % 2 == 1)
            {
                flag = false;
                break;
            }
        }
        (flag && EK() == total)? printf("possible\n"): printf("impossible\n");
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/Inkblots/p/5714704.html