bzoj4602 [Sdoi2016]齿轮

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=4602

【题解】

对于每组齿轮(u, v)连边，权值为y/x（反向边x/y）

那么直接dfs计算一遍即可。

# include <math.h>
# include <stdio.h>
# include <string.h>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10;
const int mod = 1e9+7;

# define RG register
# define ST static

int T, n, m;
int head[M], nxt[M], to[M], tot;
ld w[M];
inline void add(int u, int v, ld _w) {
    ++tot; nxt[tot] = head[u];
    head[u] = tot; to[tot] = v; w[tot] = _w;
}

bool vis[M];
ld v[M];

inline bool dfs(int x, ld c) {
    v[x] = c; vis[x] = 1;
    for (int i=head[x]; i; i=nxt[i]) {
        if(!vis[to[i]]) {
            if(dfs(to[i], c*w[i])) return 1;
        } else {
            if(fabs(v[to[i]]-c*w[i]) > 1e-8) return 1;
        }
    }
    return 0;
}
inline void sol() {
    memset(head, 0, sizeof head);
    tot = 0;
    scanf("%d%d", &n, &m);
    for (int i=1; i<=m; ++i) {
        int u, v, x, y; scanf("%d%d%d%d", &u, &v, &x, &y);
        add(u, v, (ld)y/x);
        add(v, u, (ld)x/y);
    }
    memset(vis, 0, sizeof vis);
    for (int i=1; i<=n; ++i) {
        if(vis[i]) continue;
        if(dfs(i, 1)) {
            puts("No");
            return;
        }
    }
    puts("Yes");
}

int main() {
    int T; scanf("%d", &T);
    for (int i=1; i<=T; ++i) {
        printf("Case #%d: ", i);
        sol();
    }
    return 0;
}