poj_3259

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题目链接：http://poj.org/problem?id=3259

题目大意不好解释，化成模型就是找图中是否存在负权回路，用Bellmanford就可以。

题中有说存在重边，我们要取短的那一个，其实在加边时不用管，因为在松弛操作时，长的边自然就被更新成短的边了。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<set>#include<map>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll  __int64#define N 2555int n,m;struct Node{int s,t;int val;}a[N*2+100];int k;bool BellmanFord(){int dis[N];for(int i=0;i<=n;i++)dis[i] = INF;dis[1]=0;//Init();for(int i=0;i<n;i++){int flag=0;for(int j=0;j<k;j++){int ss = a[j].s;int tt = a[j].t;int vval = a[j].val;if(dis[ss]!=INF&&dis[tt]>dis[ss]+vval){//relaxdis[tt] = dis[ss]+vval;flag=1;}}if(!flag)break;}for(int i=0;i<k;i++){if(dis[a[i].t]>dis[a[i].s]+a[i].val)//if exists negative power circuitreturn true;}return false;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//    freopen("out.txt","w",stdout);#endifint t;sf(t);while(t--){int w;k=0;memset(a,0,sizeof(struct Node)*(N*2+100));scanf("%d%d%d",&n,&m,&w);int s,t,val;for(int i=0;i<m;i++){scanf("%d%d%d",&s,&t,&val);//it's Bidirectionala[k].s = s;a[k].t = t;a[k++].val = val;a[k].s = t;a[k].t = s;a[k++].val = val;}for(int i=0;i<w;i++){scanf("%d%d%d",&s,&t,&val);//it's one waya[k].s = s;a[k].t = t;a[k++].val = -val;}if(BellmanFord())printf("YES\n");elseprintf("NO\n");}return 0;}

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