Bellman_Ford和Spfa两种方法判断负环

  Bellman_Ford是通过N-1次循环求出每个点到原点的最短路的，每次循环遍历所有的边，如果能更新就更新。前面说过为什么N-1次就行（若一个点到起点的最短路需要经过N个点，包括他自己，那么第N次循环就能确定他的最短路）。

  最短路一定是不含环的，如果存在负环就根本不存在最短路了。所以这个方法可以用来判断是否有负环，如果循环了N-1次后还能进行松弛操作，说明有负环。Bellman_Ford的复杂度是O(M*N)。

 

int bellman(){   //G条边    int i,j,k;    memset(d,INF,sizeof(d));    d[0]=0;    for(i=0;i<N-1;i++){        for(j=0;j<G;j++){            int u=e[j].u,v=e[j].v;                if(d[u]!=INF&&d[u]+w[u][v]<d[v]) d[v]=d[u]+w[u][v];        }    }    for(j=0;j<G;j++){        int u=e[j].u,v=e[j].v;            if(d[u]!=INF&&d[u]+w[u][v]<d[v]) return 1;    }    return 0;}

  因为再次循环的时候有些点的d根本没有被更新，再用这个点去更新别的点的话就浪费时间了。对Bellman_Ford改进，Spfa算法是用一个队列，每次取出一个点，更新和他连接的点，如果被更新的点不在队列中就加入队列中，这样就避免了那个问题。因为是对每个点连接的点操作，这就需要链接表了（可以用vector数组）。还需要一个数组inq判断当前这个点是否在队列里。

  如果用spfa判断是否有负环问题，其实原理也是一个点不能被更新N-1次以上，所以再开一个数组记录每个点被加入队列的次数。在网上看说是一个点被加入N次以上就说明有负环，想了半天为什么是N次以上。。。后来发现因为这是有特殊情况的，N=1，这个点被加入一次，但是没负环。。。所以只要判断这个特殊情况，后面条件写成N-1次以上也是对的。。当然为了方便，直接写成N次以上更好。。网上说这个期望时间复杂度O（2*e），这个我就不知道怎么算的了。。。

int spfa(){    queue<int> q;    memset(d,INF,sizeof(d));    memset(inq,0,sizeof(inq));    memset(cnt,0,sizeof(cnt));    d[0]=0;    inq[0]=1;    q.push(0);    while(!q.empty()){        int x=q.front();        q.pop();        inq[x]=0;        cnt[x]++;        if(cnt[x]>N) return 1;        if(d[x]==INF) continue;        for(int i=0;i<v[x].size();i++){            int y=v[x][i].y;            if(d[x]+v[x][i].t<d[y]){                d[y]=d[x]+v[x][i].t;                inq[y]=1;                q.push(y);            }        }    }    return 0;}

  Wormholes 

In the year 2163, wormholes were discovered. A wormhole is a subspace tunnel through space and time connecting two star systems. Wormholes have a few peculiar properties:

• Wormholes are one-way only.
• The time it takes to travel through a wormhole is negligible.
• A wormhole has two end points, each situated in a star system.
• A star system may have more than one wormhole end point within its boundaries.
• For some unknown reason, starting from our solar system, it is always possible to end up in any star system by following a sequence of wormholes (maybe Earth is the centre of the universe).
• Between any pair of star systems, there is at most one wormhole in either direction.
• There are no wormholes with both end points in the same star system.

All wormholes have a constant time difference between their end points. For example, a specific wormhole may cause the person travelling through it to end up 15 years in the future. Another wormhole may cause the person to end up 42 years in the past.

A brilliant physicist, living on earth, wants to use wormholes to study the Big Bang. Since warp drive has not been invented yet, it is not possible for her to travel from one star system to another one directly. Thiscan be done using wormholes, of course.

The scientist wants to reach a cycle of wormholes somewhere in the universe that causes her to end up in the past. By travelling along this cycle a lot of times, the scientist is able to go back as far in time as necessary to reach the beginning of the universe and see the Big Bang with her own eyes. Write a program to find out whether such a cycle exists.

Input 

The input file starts with a line containing the number of cases c to be analysed. Each case starts with a line with two numbersn andm . These indicate the number of star systems () and the number of wormholes () . The star systems are numbered from 0 (our solar system) throughn-1 . For each wormhole a line containing three integer numbersx,y andt is given. These numbers indicate that this wormhole allows someone to travel from the star system numberedx to the star system numberedy, thereby ending upt () years in the future.

Output 

The output consists of c lines, one line for each case, containing the wordpossible if it is indeed possible to go back in time indefinitely, ornot possible if this is not possible with the given set of star systems and wormholes.

Sample Input 

23 30 1 10001 2 152 1 -424 40 1 101 2 202 3 303 0 -60

Sample Output 

possiblenot possible

#include<cstring>#include<cstdio>#include<iostream>#include<cmath>#include<algorithm>#include<queue>#include<map>#include<vector>#define INF 0x3f3f3f3f#define MAXN 1010#define pii pair<int,int>using namespace std;int N,M,G,w[MAXN][MAXN],d[MAXN],inq[MAXN],cnt[MAXN];struct edge{    int y,t;    edge(int a,int b){        y=a;        t=b;    }};vector<edge> v[MAXN];struct Bellman_edge{    int u,v,t;    Bellman_edge(){}    Bellman_edge(int a,int b,int d){        u=a;        v=b;        t=d;    }}e[2*MAXN];int bellman(){   //G条边    int i,j,k;    memset(d,INF,sizeof(d));    d[0]=0;    for(i=0;i<N-1;i++){        for(j=0;j<G;j++){            int u=e[j].u,v=e[j].v;                if(d[u]!=INF&&d[u]+w[u][v]<d[v]) d[v]=d[u]+w[u][v];        }    }    for(j=0;j<G;j++){        int u=e[j].u,v=e[j].v;            if(d[u]!=INF&&d[u]+w[u][v]<d[v]) return 1;    }    return 0;}int spfa(){    queue<int> q;    memset(d,INF,sizeof(d));    memset(inq,0,sizeof(inq));    memset(cnt,0,sizeof(cnt));    d[0]=0;    inq[0]=1;    q.push(0);    while(!q.empty()){        int x=q.front();        q.pop();        inq[x]=0;        cnt[x]++;        if(cnt[x]>N) return 1;        if(d[x]==INF) continue;        for(int i=0;i<v[x].size();i++){            int y=v[x][i].y;            if(d[x]+v[x][i].t<d[y]){                d[y]=d[x]+v[x][i].t;                inq[y]=1;                q.push(y);            }        }    }    return 0;}int main(){    freopen("in.txt","r",stdin);    int T;    scanf("%d",&T);    while(T--){        scanf("%d%d",&N,&M);        memset(w,INF,sizeof(w));        int i,a,b,t;        G=0;        for(i=0;i<N;i++) v[i].clear();        for(i=0;i<N;i++) w[i][i]=0;        for(i=0;i<M;i++){            scanf("%d%d%d",&a,&b,&t);            w[a][b]=t;                      //bellman            e[G++]=Bellman_edge(a,b,t);            v[a].push_back(edge(b,t));      //spfa        }        if(bellman()) printf("possible\n");        else printf("not possible\n");    }    return 0;}