poj 3259spfa()判断是否存在负环

http://poj.org/problem?id=3259

Wormholes

Time Limit: 2000MSMemory Limit: 65536KTotal Submissions: 22876Accepted: 8144

Description

While exploring his many farms, Farmer John has discovered anumber of amazing wormholes. A wormhole is very peculiar because itis a one-way path that delivers you to its destination at a timethat is BEFORE you entered the wormhole! Each of FJ's farmscomprises N (1≤ N ≤ 500) fieldsconveniently numbered1..N, M (1≤ M ≤ 2500) paths,and W (1≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do thefollowing: start at some field, travel through some paths andwormholes, and return to the starting field a time before hisinitial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he willsupply you with complete mapsto F (1≤ F ≤ 5) of his farms. Nopaths will take longer than 10,000 seconds to travel and nowormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A singleinteger, F. F farmdescriptions follow. 
Line 1 of each farm: Three space-separated integersrespectively: N, M,and W 
Lines 2..M+1 of each farm: Three space-separated numbers(S, E, T)that describe, respectively: a bidirectional pathbetween S and E thatrequires T seconds totraverse. Two fields might be connected by more than onepath. 
Lines M+2..M+W+1 of eachfarm: Three space-separated numbers(S, E, T)that describe, respectively: A one way pathfrom S to E thatalso moves the travelerback T seconds.

Output

Lines 1..F: For each farm, output "YES" ifFJ can achieve his goal, otherwise output "NO" (do not include thequotes).

Sample Input

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

Hint

For farm 1, FJ cannot travel back intime. 
For farm 2, FJ could travel back in time by the cycle1->2->3->1, arriving back at his starting location 1second before he leaves. He could start from anywhere on the cycleto accomplish this.

Source

USACO 2006 December Gold

//题解：图论算法。

//ac

// I'm theTopcoder

//C

#include

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//C++

#include

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#include

using namespacestd;

//*************************OUTPUT*************************

#ifdefWIN32

#define INT64"%I64d"

#define UINT64"%I64u"

#else

#define INT64"%lld"

#define UINT64"%llu"

#endif

//**************************CONSTANT***********************

#define INF0x3f3f3f3f

#define eps1e-8

#define PIacos(-1.)

#define PI2asin (1.);

typedef longlong LL;

//typedef__int64 LL;   //codeforces

typedefunsigned int ui;

typedefunsigned long long ui64;

#define MPmake_pair

typedef vectorVI;

typedef pairPII;

#define pbpush_back

#define mpmake_pair

//***************************SENTENCE************************

#define CL(a,b)memset (a, b, sizeof (a))

#definesqr(a,b) sqrt ((double)(a)*(a) +(double)(b)*(b))

#definesqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) +(double)(c)*(c))

//****************************FUNCTION************************

template doubleDIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y);}

template inlineT INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; returnlen; }

template inlineT square(T va, T vb) { return va * va + vb * vb;}

// aply for thememory of the stack

//#pragmacomment (linker,"/STACK:1024000000,1024000000")

//end

const int maxn= 1000000+100;

structnode{

   int to;

   int next;

   long long weight;

};

nodeedge[maxn],edge1[maxn];//保存边的起点和终点

intn,m;

long longval;

inttot,tot1;

intsrc;//起点

inthead[maxn],head1[maxn];

intvisit[maxn],visit1[maxn];

long longdis[maxn],dis1[maxn];

long longans[maxn],ans2[maxn];

long longMAXTIME=-INF;

intsum1[maxn],sum11[maxn];;

void add(inta,int b,long long c){

   edge[tot].to=b;

   edge[tot].weight=c;

   edge[tot].next=head[a];

   head[a]=tot++;

}

void add1(inta,int b,long long c){

   edge1[tot1].to=b;

   edge1[tot1].weight=c;

   edge1[tot1].next=head1[a];

   head1[a]=tot1++;

}

boolspfa(){

   //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 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;

       for(inti=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;

}

intmain(){

   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(inti=1;i<=n;i++){

         head[i]=-1;

         head1[i]=-1;

      }

       for(inti=1;i<=m;i++){

         scanf("%d%d%d",&a,&b,&c);

         add(a,b,c);

         add(b,a,c);

         //add1(b,a,c);

      }

       for(inti=1;i<=w;i++){

         scanf("%d%d%d",&a,&b,&c);

         add(a,b,-c);

      }

      src=1;

      if(spfa()){

         printf("YES\n");

      }

       elseprintf("NO\n");

   }

   return 0;

}