hdu6073 Matching In Multiplication 分析+拓扑序

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 787 Accepted Submission(s): 222

Problem Description

In the mathematical discipline of graph theory, a bipartite graph is a graph whose vertices can be divided into two disjoint sets

Input

The first line of the input contains an integer

.

It is guaranteed that each graph has at least one perfect matchings, and there are at most one edge between every pair of vertex.

Output

For each test case, print a single line containing an integer, denoting the weight of the given graph. Since the answer may be very large, please print the answer modulo

Sample Input

1 2 2 1 1 4 1 4 2 3

Sample Output

/**
题目：hdu6073 Matching In Multiplication
链接：http://acm.hdu.edu.cn/showproblem.php?pid=6073
题意：

思路：
首先如果一个点的度数为1，那么它的匹配方案是固定的，继而我们可以去掉这一对点。通过拓扑我们可以不断去掉所有度数为1的点。

那么剩下的图中左右各有m个点，每个点度数都不小于2，且左边每个点度数都是2，而右侧总度数是2m，因此右侧只能是每个点度数都是2。

这说明这个图每个连通块是个环，在环上间隔着取即可，一共两种方案。

时间复杂度O(n)。

*/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long LL;
#define lson L,m,rt<<1
typedef pair<int,int> P;
#define rson m+1,R,rt<<1|1
const int mod = 998244353;
const double eps = 1e-4;
const int N = 6e5+10;
int cnt[N], vis[N];
int a[N], an, n;
LL ans;
vector<P>G[N];
queue<int> q;
int now;
void solve(int r,int f,LL &ansl,LL &ansr,int step)
{
    for(int i = 0; i < (int)G[r].size(); i++){
        if(G[r][i].first!=f&&(vis[G[r][i].first]==0||G[r][i].first==now)){
            vis[G[r][i].first] = 1;
            if(step%2==0){
                ansl = ansl*G[r][i].second%mod;
            }else
            {
                ansr = ansr*G[r][i].second%mod;
            }
            if(G[r][i].first==now){///回到起点。
                return ;
            }else
                return solve(G[r][i].first,r,ansl,ansr,step+1);
        }
    }
}
/*
void input()
{
    for(int i = 1; i <= n; i+=2){
        G[i].push_back(P(i+n,1));
        G[i].push_back(P(i+n+1,1));
        G[i+1].push_back(P(i+n,1));
        G[i+1].push_back(P(i+n+1,1));
        G[i+n].push_back(P(i,1));
        G[i+n+1].push_back(P(i,1));
        G[i+n].push_back(P(1+i,1));
        G[i+n+1].push_back(P(1+i,1));
        cnt[i+n]+=2;
        cnt[i+n+1]+=2;
    }
}*/
int main()
{
    //freopen("C:\\Users\\accqx\\Desktop\\in.txt","r",stdin);
    int T;
    cin>>T;
    int u1, w1, u2, w2;
    while(T--)
    {
        scanf("%d",&n);
        memset(cnt, 0, sizeof cnt);
        memset(vis, 0, sizeof vis);
        for(int i = 1; i <= 2*n; i++) G[i].clear();
        //input();
        for(int i = 1; i <= n; i++){
            scanf("%d%d%d%d",&u1,&w1,&u2,&w2);
            G[i].push_back(P(u1+n,w1));
            G[i].push_back(P(u2+n,w2));
            G[u1+n].push_back(P(i,w1));
            G[u2+n].push_back(P(i,w2));
            cnt[u1+n]++;
            cnt[u2+n]++;
        }
        ans = 1;
        while(!q.empty()) q.pop();
        for(int i = n+1; i <= n*2; i++){
            if(cnt[i]==1){
                q.push(i);
            }
        }
        while(!q.empty()){
            int r = q.front();
            q.pop();
            int len = G[r].size();
            int pos;
            for(int i = 0; i < len; i++){
                if(vis[G[r][i].first]==0){
                    vis[G[r][i].first] = 1;
                    ans = ans*G[r][i].second%mod;
                    pos = G[r][i].first;
                    break;
                }
            }

            len = G[pos].size();
            for(int i = 0; i < len; i++){
                if(G[pos][i].first!=r){
                    cnt[G[pos][i].first]--;
                    if(cnt[G[pos][i].first]==1){
                        q.push(G[pos][i].first);
                    }
                }
            }
        }
        LL ansl, ansr;
        for(int i = 1; i <= n; i++){
            if(vis[i]==0){
                now = i;
                vis[i] = 1;
                ansl = ansr = 1;
                solve(i,-1,ansl,ansr,0);
                ans = ans*(ansl+ansr)%mod;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}