【题解】LuoGu1993：小 K 的农场

原题传送门
差分约束

• 操作1： $addedge(b,a,c)$
• 操作2： $addedge(a,b,-c)$
• 操作3： $addedge(a,b,0),addedge(b,a,0)$

用 $spfa$跑个最长路

Code：

#include <bits/stdc++.h>
#define maxn 100010
using namespace std;
struct Edge{
	int to, next, len;
}edge[maxn << 1];
queue <int> q;
int dis[maxn], vis[maxn], num, head[maxn], n, m, tot[maxn];

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = 1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

void addedge(int x, int y, int z){ edge[++num] = (Edge){y, head[x], z}, head[x] = num; }

int main(){
	freopen("farm.in", "r", stdin);
	freopen("farm.out", "w", stdout);
	n = read(), m = read();
	for (int i = 1; i <= n; ++i) addedge(0, i, 0);
	for (int i = 1; i <= m; ++i){
		int opt = read(), x = read(), y = read();
		if (opt == 1) addedge(y, x, read());
		else if (opt == 2) addedge(x, y, -read());
		else addedge(x, y, 0), addedge(y, x, 0);
	}
	for (int i = 1; i <= n; ++i) dis[i] = -1e9;
	vis[0] = 1;
	q.push(0);
	int flag = 0;
	while (!q.empty()){
		int u = q.front(); q.pop();
		vis[u] = 0;
		for (int i = head[u]; i; i = edge[i].next){
			int v = edge[i].to;
			if (dis[v] < dis[u] + edge[i].len){
				dis[v] = dis[u] + edge[i].len;
				if ((++tot[v]) > n){ flag = 1; break; }
				if (!vis[v]) vis[v] = 1, q.push(v);
			}
		}
		if (flag) break;
	}
	if (flag) puts("No"); else puts("Yes");
	return 0;
}