luogu P1993 小K的农场

刚看这道题题面时并没有觉得这是道图论题。后来仔细想了想发现spfa可以做（遇图不决spfa）在输入的时候加几个判定就完事了。

上代码

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 50005;
int n, m;
inline int read() {
    int a=0;
    char x=getchar();
    while(x<'0'||x>'9')x=getchar();
    while(x>='0'&&x<='9')a=(a<<3)+(a<<1)+x-48,x=getchar();
    return a;
}
struct edge {
    int to, nxt, w;
} e[maxn];
int dis[maxn], cnt, tot[maxn], dep[maxn];
bool vis[maxn];
inline void add(int u, int v, int W) {
    e[++cnt].to = v;
    e[cnt].nxt = dep[u];
    dep[u] = cnt;
    e[cnt].w = W;
}
inline bool spfa(int u) {
    vis[u] = 1;
    for(int i = dep[u]; i; i = e[i].nxt)
        if(dis[e[i].to] < dis[u] + e[i].w) {
            dis[e[i].to] = dis[u] + e[i].w;
            if(vis[e[i].to])
                return 0;
            if(!spfa(e[i].to))
                return 0;
        }
    vis[u]= 0;
    return 1;
}
int num, a, b, c;
int main() {
    n = read();
    m = read();
    while(m--) {
        num = read();
        a = read();
        b = read();
        if(num == 1) {
            c = read();
            add(b,a,c);
        } else if(num == 2) {
            c = read();
            add(a,b,-c);
        } else if(num == 3) {
            add(a,b,0);
            add(b,a,0);
        }
    }
    for(int i = 1; i <= n; i++) {
        add(0,i,0);
        dis[i] = -23333333;
    }
    if(!spfa(0))
        printf("No");
    else
        printf("Yes");
    return 0;
}