HDU-6073 Matching In Multiplication（拓扑+dfs）

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传送门：HDU-6073

xjb拓扑一下把只有一条边的点去掉，然后dfs一下就行，中间TLE到怀疑人生，重写一遍就AC了，现在已经不想知道到底哪有问题了

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int mod = 998244353;const int MX = 300000 + 10;struct Edge {    int nxt, v, w;    bool vis;} E[MX * 4];int T, n, head[MX * 2], tot, IN[MX * 2], vis[MX * 2];LL val[2], ans;void add(int u, int v, int w) {    E[tot].v = v;    E[tot].w = w;    E[tot].vis = 0;    E[tot].nxt = head[u];    head[u] = tot++;    IN[v]++;}void dfs(int u, int idx) {    vis[u] = 1;    for (int i = head[u]; i; i = E[i].nxt) {        if (E[i].vis)    continue;        E[i].vis = E[i ^ 1].vis = 1;        val[idx] = val[idx] * E[i].w % mod;        dfs(E[i].v, 1 ^ idx);    }}void top_sort() {    queue<int> q;    for (int v = n + 1; v <= n + n; v++) if (IN[v] == 1)  q.push(v);    while (!q.empty()) {        int v = q.front();        q.pop();        vis[v] = 1;        for (int i = head[v]; i; i = E[i].nxt) {            if (E[i].vis)    continue;            E[i].vis = E[i ^ 1].vis = 1;            vis[ E[i].v ] = 1;            ans = ans * E[i].w % mod;            for (int j = head[ E[i].v ]; j; j = E[j].nxt) {                E[j].vis = E[j ^ 1].vis = 1;                IN[E[j].v]--;                if (IN[ E[j].v ] == 1) q.push(E[j].v);            }        }    }}void solve() {    ans = 1;    top_sort();    for (int u = 1; u <= n; u++) {        if (vis[u]) continue;        val[0] = val[1] = 1;        dfs(u, 0);        ans = ans * (val[0] + val[1]) % mod;    }}int main() {    scanf("%d", &T);    while (T--) {        scanf("%d", &n);        memset(head, 0, sizeof(head));        memset(IN, 0, sizeof(IN));        memset(vis, 0, sizeof(vis));        tot = 0;        for (int u = 1, v, w; u <= n; u++)            for (int i = 0; i < 2; i++) {                scanf("%d %d", &v, &w);                add(u, v + n, w);                add(v + n, u, w);            }        solve();        printf("%lld\n", ans);    }}

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