2017多校第4场 HDU 6073 Matching In Multiplication 拓扑排序，思维，DFS

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6073

题意：给个图，求每个完全匹配的权值的积，最后再求和

解法：首先用拓扑排序处理一下度为1的边，这些边对答案的贡献就是它的权值，不难想到答案就是ans1*sigma(每个连通块的贡献)，所谓连通块就是指删掉度为1的点，剩下的联通块的方案数，所以在拓扑的过程中直接乘以度为1的边的权值，之后对每个环DFS，得到当前这个环贡献的答案，由于是匹配只有两种答案，交替走就可以了，这个用DFS可以轻松搬办到。

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int maxn = 300010;const int mod = 998244353;struct FastIO{    static const int S = 1310720;    int wpos;    char wbuf[S];    FastIO() : wpos(0) {}    inline int xchar()    {        static char buf[S];        static int len = 0, pos = 0;        if (pos == len)            pos = 0, len = fread(buf, 1, S, stdin);        if (pos == len) return -1;        return buf[pos ++];    }    inline int xuint()    {        int c = xchar(), x = 0;        while (c <= 32) c = xchar();        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';        return x;    }    inline int xint()    {        int s = 1, c = xchar(), x = 0;        while (c <= 32) c = xchar();        if (c == '-') s = -1, c = xchar();        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';        return x * s;    }    inline void xstring(char *s)    {        int c = xchar();        while (c <= 32) c = xchar();        for (; c > 32; c = xchar()) * s++ = c;        *s = 0;    }    inline void wchar(int x)    {        if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;        wbuf[wpos ++] = x;    }    inline void wint(LL x)    {        if (x < 0) wchar('-'), x = -x;        char s[24];        int n = 0;        while (x || !n) s[n ++] = '0' + x % 10, x /= 10;        while (n--) wchar(s[n]);    }    inline void wstring(const char *s)    {        while (*s) wchar(*s++);    }    ~FastIO()    {        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;    }} io;struct edge{    int to,w,next;    bool mark;}E[maxn*4];int T, n, head[maxn*2], edgecnt, du[maxn*2], used[maxn*2];LL temp[2];void init(){    memset(head,-1,sizeof(head));    edgecnt=0;}void add(int u, int v, int w){    E[edgecnt].to=v,E[edgecnt].w=w,E[edgecnt].next=head[u],E[edgecnt].mark=0,head[u]=edgecnt++;}void dfs(int x, int id){    used[x]=1;    for(int i=head[x];~i;i=E[i].next){        if(E[i].mark) continue;        E[i].mark=E[i^1].mark=1;        temp[id]*=E[i].w;        temp[id]%=mod;        dfs(E[i].to, 1-id);    }}int main(){    T = io.xint();    while(T--){        init();        n = io.xint();        memset(du, 0, sizeof(du));        memset(used, 0, sizeof(used));        int u,v,w;        for(u=1; u<=n; u++){            for(int i=1; i<=2; i++){                //scanf("%d%d",&v,&w);                v = io.xint();                w = io.xint();                add(u,v+n,w);                add(v+n,u,w);                du[u]++;                du[v+n]++;            }        }        LL ans=1;        queue <int> q;        for(u=n+1; u<=n+n; u++){            if(du[u]==1){                q.push(u);            }        }        while(q.size())        {            u=q.front(); q.pop();            used[u]=1;            for(int i=head[u]; ~i; i=E[i].next){                if(E[i].mark) continue;                E[i].mark=E[i^1].mark=1;                used[E[i].to]=1;                ans*=E[i].w;                ans%=mod;                for(int j=head[E[i].to]; ~j; j=E[j].next){                    E[j].mark=E[j^1].mark=1;                    du[E[j].to]--;                    if(du[E[j].to]==1) q.push(E[j].to);                }            }        }        for(int i=1; i<=n; i++){            if(used[i]) continue;            temp[0]=temp[1]=1;            dfs(i,0);            ans=(ans%mod*(temp[0]+temp[1])%mod)%mod;        }        ans %= mod;        printf("%lld\n", ans);    }    return 0;}