P2294 [HNOI2005] cunning businessman (differential constraints)

P2294 [HNOI2005] cunning businessman

For each $ (x, y, w) $, even edge $ (x-1, y, w), (y, x-1, -w) $, $ y represents earnings before months ago than $ $ x-1 $ months of big gains $ w $

Such subject was asked whether there is conversion of FIG nonzero rings present

So we look for positive ring (longest path) and negative loop (shortest) just fine

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In fact, you find positive and negative loop ring are equivalent

So you only require one on the line

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 105
int T,n,m,a[N][N],f1[N],f2[N],inf;
bool vis1[N],vis2[N],ok;
void spfa1(int x){
    if(vis1[x]) ok=1;
    if(ok) return ;
    vis1[x]=1;
    for(int i=0;i<=n&&!ok;++i)
        if(a[x][i]!=inf)
            if(f1[i]>f1[x]+a[x][i])
                f1[i]=f1[x]+a[x][i],spfa1(i);
    vis1[x]=0;
}
void spfa2(int x){
    if(vis2[x]) ok=1;
    if(ok) return ;
    vis2[x]=1;
    for(int i=0;i<=n&&!ok;++i)
        if(a[x][i]!=inf)
            if(f2[i]<f2[x]+a[x][i])
                f2[i]=f2[x]+a[x][i],spfa2(i);
    vis2[x]=0;
}
int main(){
    scanf("%d",&T);
    while(T--){
        memset(a,127,sizeof(a)); inf=a[0][0];
        memset(f1,127,sizeof(f1));
        memset(f2,128,sizeof(f2));
        scanf("%d%d",&n,&m); ok=0; f1[0]=f2[0]=0;
        for(int i=1,u,v,w;i<=m;++i){
            scanf("%d%d%d",&u,&v,&w);
            a[u-1][v]=w;
            a[v][u-1]=-w;
        }
        for(int i=0;i<=n;++i) spfa1(i),spfa2(i);//自选一种
        puts(ok?"false":"true");
    }return 0;
}