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。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
# 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; }