HDU 6073 Matching In Multiplication（拓扑排序）

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1127 Accepted Submission(s): 325

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

【题意】给你一个二分图，每一集合里的点数量都为n，且其中一个集合里每个点的度数都为2.然后对于每一种完美匹配，算出边权值的乘积，然后再将每一种匹配的乘积加起来，输出最后结果。

【分析】U集合里的点度数都为二，V集合里的点度数未知，但加起来肯定为2*n，对于V集合里度数为1的点，它的匹配对象是固定的，所以我们用拓扑排序将度数为1的点全部挖出，算出乘积res。然后对于剩下的图，假设总节点为2*m，则V集合总度数为2*m，由于此时V集合里已经，没有度数为1的点，所以V集合里点的度数都为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。比如对于两个联通块，他俩的两种方案乘积分别是(a1,a2),(b1,b2)，则答案为a1*b1+a1*b2+a2*b1+a2*b2,化简后为(a1+a2)*(b1+b2),最后在乘以res即可。

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define qwer 2e18
using namespace std;
typedef long long ll;
const int N = 6e5+50;
const int M = 16000009;
const int mod = 998244353;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
int n,s;
int vis[N],in[N];
ll ans[2];
vector<pii>edg[N];
ll topSort(){
    queue<int>q;
    ll ret=1;
    for(int i=n+1;i<=n+n;i++){
        if(in[i]==1){
            q.push(i);
            vis[i]=1;
        }
    }
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=0;i<edg[u].size();i++){
            int v=edg[u][i].first;
            if(vis[v])continue;
            if((--in[v])==1)q.push(v),vis[v]=1;
            if(u>n)ret=(ret*1LL*edg[u][i].second)%mod;
        }
    }
    return ret;
}
void dfs(int u,int ty,int fa){
    vis[u]=1;
    for(int i=0;i<edg[u].size();i++){
        int v=edg[u][i].first;
        if(v==s&&v!=fa)ans[ty]=(ans[ty]*1LL*edg[u][i].second)%mod;
        if(vis[v])continue;
        ans[ty]=(ans[ty]*1LL*edg[u][i].second)%mod;
        dfs(v,ty^1,u);
    }
}
int main(){
    //freopen("de.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i=0;i<N;i++)vis[i]=in[i]=0,edg[i].clear();
        for(int i=1,v1,w1,v2,w2;i<=n;i++){
            scanf("%d%d%d%d",&v1,&w1,&v2,&w2);
            v1+=n;v2+=n;
            edg[i].pb(mp(v1,w1));
            edg[v1].pb(mp(i,w1));
            edg[i].pb(mp(v2,w2));
            edg[v2].pb(mp(i,w2));
            in[i]+=2;
            in[v1]++;in[v2]++;
        }
        ll anss=topSort();
        for(s=1;s<=n;s++){
            if(!vis[s]){
                ans[0]=ans[1]=1;
                dfs(s,0,0);
                anss=anss*((ans[0]+ans[1])%mod)%mod;
            }
        }
        printf("%lld\n",anss);
    }
    return 0;
}