差分约束/并查集——BZOJ1202/Luogu2294 [HNOI2005]狡猾的商人

题面：Luogu2294 BZOJ1202
带权并查集直接做
路径压缩的时候把并查集的权值算好，最后判断矛盾是只要前缀减一下就好了
还有一种做法是差分约束，暴力按照题意建边然后spfa判负环即可
luogu实测两者的速度都在130ms左右，差距不大
并查集：

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<iostream>#include<cstdlib>#include<set>#include<map>#include<string>#include<ctime>#include<queue>#include<vector>using namespace std;int n,m,fa[100001],v[100001];inline int getfather(int x){    if(fa[x]==x)return x;    int t=getfather(fa[x]);    v[x]+=v[fa[x]];return fa[x]=t;}int main(){    int T;scanf("%d",&T);    while(T--){        bool flag=0;        scanf("%d%d",&n,&m);        for(int i=0;i<=n;i++)fa[i]=i,v[i]=0;        for(int i=1;i<=m;i++){            int x,y,z;scanf("%d%d%d",&x,&y,&z);x--;            int fx=getfather(x),fy=getfather(y);            if(fx!=fy){                fa[fx]=fy;v[fx]=v[y]-v[x]-z;            }else if(v[y]-v[x]!=z){flag=1;break;}        }        puts(flag?"false":"true");    }    return 0;}

差分约束：

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<iostream>#include<cstdlib>#include<set>#include<map>#include<string>#include<ctime>#include<queue>#include<vector>using namespace std;bool vis[1001];int n,m,dist[2001];int nedge=0,p[2001],c[2001],nex[2001],head[2001];inline void addedge(int x,int y,int z){    p[++nedge]=y;c[nedge]=z;    nex[nedge]=head[x];head[x]=nedge;}inline bool spfa(int x){    vis[x]=1;    for(int k=head[x];k;k=nex[k])if(dist[p[k]]>dist[x]+c[k]){        dist[p[k]]=dist[x]+c[k];        if(vis[p[k]]||!spfa(p[k]))return 0;    }    vis[x]=0;return 1;}int main(){    int T;scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&m);        nedge=0;memset(p,0,sizeof p);memset(c,0,sizeof c);        memset(nex,0,sizeof nex);memset(head,0,sizeof head);        for(int i=1;i<=m;i++){            int x,y,z;scanf("%d%d%d",&x,&y,&z);            addedge(x-1,y,z);addedge(y,x-1,-z);        }        for(int i=0;i<=n;i++)dist[i]=1e9;        memset(vis,0,sizeof vis);dist[0]=0;        puts(spfa(0)?"true":"false");    }    return 0;}