BZOJ3436_小K的农场_KEY

题目传送门

差分约束基础，对于每种关系建不同的边，求是否有负环。

code：

/**************************************************************
    Problem: 3436
    User: yekehe
    Language: C++
    Result: Accepted
    Time:252 ms
    Memory:10916 kb
****************************************************************/
 
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
 
char tc()
{
    static char fl[10000000],*A=fl,*B=fl;
    return A==B&&(B=(A=fl)+fread(fl,1,10000000,stdin),A==B)?EOF:*A++;
}
 
int read()
{
    char c;while(c=getchar(),(c<'0' || c>'9') && c!='-');
    int x=0,y=1;c=='-'?y=-1:x=c-'0';
    while(c=getchar(),c>='0' && c<='9')x=x*10+c-'0';return x*y;
}
 
const int MAXN=10005;
 
int N,M;
int o,x,y,z;
int dist[MAXN],vis[MAXN];
bool flag;
vector < pair<int, int> > a[MAXN];
 
void cf_spfa(int x)
{
    int to,val;
        for(int i=0;i<a[x].size();i++){
            to=a[x][i].first;val=a[x][i].second;
            if(dist[x]+val<dist[to]){
                if(vis[to]){
                    flag=true;
                    return ;
                }
                else {
                    dist[to]=dist[x]+val;
                    vis[to]=1;
                    cf_spfa(to);
                }
            }
        }
    vis[x]=false;
}//SPFA判负环
 
int main()
{
//  freopen("x.txt","r",stdin);
    N=read(),M=read();
        for(int i=1; i<=M; i++){
            o=read();
            if(o==1) {x=read(),y=read(),z=read();a[y].push_back(make_pair(x,-z));}
            if(o==2) {x=read(),y=read(),z=read();a[x].push_back(make_pair(y,z));}
            if(o==3) {x=read(),y=read();a[x].push_back(make_pair(y,0)),a[y].push_back(make_pair(x,0));}
        }//建边
        for(int i=1; i<=N; i++) {memset(dist,0,sizeof(dist));cf_spfa(i);if(flag)break;}
    if(flag)puts("No");
    else puts("Yes");
    return 0;
}