[luogu1993] 小K的农场

传送门

Solution

算是一道差分约束的模板题了。

将符号统一一下，如果是小于号就求负环，否则求正环

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 10005

struct edge{
    int v,w,next;
}G[MAXN<<2];
int head[MAXN];
int vis[MAXN];
int dis[MAXN];
int N,M,tot = 0;

inline int read(){
    int num = 0;char ch = getchar();
    while(ch>'9'||ch<'0')ch = getchar();
    while(ch>='0'&&ch<='9')num = num*10+ch-'0',ch = getchar();
    return num;
}

inline void add(int u,int v,int w){
    G[++tot].v = v;G[tot].w = w;G[tot].next = head[u];head[u] = tot;
}

bool dfs(int u){
    vis[u] = 1;
    for(register int i=head[u];i!=-1;i=G[i].next){
        int v = G[i].v;
        if(dis[v]>dis[u]+G[i].w){
            dis[v] = dis[u] + G[i].w;
            if(vis[v]||dfs(v))return true;
        }
    }
    vis[u] = 0;
    return false;
}
int main(){

    N = read();M = read();
    std::memset(head,-1,sizeof(head));
    std::memset(vis,0,sizeof(vis));

    int opt,u,v,w;
    for(register int i=1;i<=M;++i){
        opt = read();u = read();v = read();
        if(opt==3){
            add(u,v,0);add(v,u,0);
        }
        else{
            w = read();
            if(opt==1)add(u,v,-w);
            else add(v,u,w);
        }
    }

    for(register int i=1;i<=N;++i)add(0,i,0);
    for(register int i=1;i<=N;++i)dis[i] = MAXN;
    dis[0] = 0;
    if(dfs(0))puts("No");
    else puts("Yes");

    return 0;
}