poj_3259

这道题目是求最短路图中是否存在负权回路，我们用bellman_ford即可。做此题时用了下链式向前星，因为之前没怎么用过。

#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<map>//#define ONLINE_JUDGE#define eps 1e-10#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))#define ll __int64const 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;int n,m,k;int s;#define N 8080struct Edge{    int to,next;    int w;}e[10010];int head[550];void addEdge(int u,int v,int w){    //链式向前星    e[s].to = v;    e[s].w = w;    e[s].next = head[u];    head[u] = s++;}bool bellman_Ford(){    int dis[550];    for(int i=1;i<=n;i++)        dis[i] = 0x3f3f3f3f;    dis[1] = 0;    for(int i=1;i<n;i++){        for(int j=1;j<=n;j++){  //遍历所有的点            int p = head[j];            for(;~p;p=e[p].next)//对于每个点，遍历所有它拥有的边  ~p==(p!=-1)                if(dis[e[p].to]>dis[j]+e[p].w)                    dis[e[p].to] = dis[j]+e[p].w;        }    }    for(int i=1;i<n;i++){            for(int j=1;j<=n;j++){  //寻找是否存在负权回路                int p = head[j];                for(;~p;p=e[p].next)                    if(dis[e[p].to]>dis[j]+e[p].w)                        return true;            }        }    return false;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endif    int t;    sf(t);    while(t--){        scanf("%d%d%d",&n,&m,&k);        int u,v,w;        memset(head,-1,sizeof head);        s=0;        for(int i=0;i<m;i++){            scanf("%d%d%d",&u,&v,&w);            addEdge(u,v,w);            addEdge(v,u,w);        }        for(int i=0;i<k;i++){//注意题目中所述的k条边是单项边            scanf("%d%d%d",&u,&v,&w);            addEdge(u,v,-w);        }        if(bellman_Ford())            printf("YES\n");        else            printf("NO\n");    }return 0;}