POJ 3295 spfa判断是否存在负权回路

http://blog.sina.com.cn/s/blog_99ca2df50101jtk3.html

这里前几天刚刚写过就不写了。

题意：判断是否存在负权回路。用spfa.

//ac

// I'm the Topcoder
//C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
//C++
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cctype>
#include <stack>
#include <string>
#include <list>
#include <queue>
#include <map>
#include <vector>
#include <deque>
#include <set>
using namespace std;

//*************************OUTPUT*************************
#ifdef WIN32
#define INT64 "%I64d"
#define UINT64 "%I64u"
#else
#define INT64 "%lld"
#define UINT64 "%llu"
#endif

//**************************CONSTANT***********************
#define INF 0x3f3f3f3f
#define eps 1e-8
#define PI acos(-1.)
#define PI2 asin (1.);
typedef long long LL;
//typedef __int64 LL;   //codeforces
typedef unsigned int ui;
typedef unsigned long long ui64;
#define MP make_pair
typedef vector<int> VI;
typedef pair<int, int> PII;
#define pb push_back
#define mp make_pair

//***************************SENTENCE************************
#define CL(a,b) memset (a, b, sizeof (a))
#define sqr(a,b) sqrt ((double)(a)*(a) + (double)(b)*(b))
#define sqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) + (double)(c)*(c))

//****************************FUNCTION************************
template <typename T> double DIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y); }
template <class T> inline T INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; return len; }
template <typename T> inline T square(T va, T vb) { return va * va + vb * vb; }

// aply for the memory of the stack
//#pragma comment (linker, "/STACK:1024000000,1024000000")
//end

const int maxn = 1000000+100;
struct node{
    int to;
    int next;
    long long weight;
};
node edge[maxn],edge1[maxn];//保存边的起点和终点
int n,m;
long long val;
int tot,tot1;
int src;//起点
int head[maxn],head1[maxn];
int visit[maxn],visit1[maxn];
long long dis[maxn],dis1[maxn];
long long ans[maxn],ans2[maxn];
long long MAXTIME=-INF;
int sum1[maxn],sum11[maxn];;

void add(int a,int b,long long c){
    edge[tot].to=b;
    edge[tot].weight=c;
    edge[tot].next=head[a];
    head[a]=tot++;
}

void add1(int a,int b,long long c){
    edge1[tot1].to=b;
    edge1[tot1].weight=c;
    edge1[tot1].next=head1[a];
    head1[a]=tot1++;
}

bool spfa(){
    //初始化
    for(int i=1;i<=n;i++){
        dis[i]=INF;
        visit[i]=0;//访问标记
        sum1[i]=0;
    }
    dis[src]=0; visit[src]=1;
    int u;
    int v;
    queue<int> Q;//优先队列
    Q.push(src);
    while(!Q.empty()){
        u=Q.front();
        Q.pop();
        visit[u]=0;
        if(sum1[u]>n) return true;//一个点入队列次数超过n次则表示存在负权回路
        for(int i=head[u];i!=-1;i=edge[i].next){
            v=edge[i].to;
            if(dis[v]>dis[u]+edge[i].weight){
                dis[v]=dis[u]+edge[i].weight;
                if(!visit[v]){
                    Q.push(v);
                    visit[v]=1;
                    sum1[v]++;
                }
            }
        }
    }
    return false;
}



int main(){
    int a,b,c,w;
    int t;
    scanf("%d",&t);
    while( t--){
        tot=tot1=0;//边的条数
        scanf("%d%d%d",&n,&m,&w);//w个洞
        for(int i=1;i<=n;i++){
            head[i]=-1;
            head1[i]=-1;
        }
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
            //add1(b,a,c);
        }
        for(int i=1;i<=w;i++){
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,-c);
        }
        src=1;
        if(spfa()){
            printf("YES\n");
        }
        else printf("NO\n");
    }
    return 0;
}