uva558

来源：互联网发布：c语言更改define 编辑：程序博客网时间：2024/04/29 18:34

题目大意：
n个点，m条边，边给出的信息为：两个端点编号，以及权值，权值可为负数。判断图中是否存在负环。

思路：
bellmanford判断负环
spaf bfs判断负环

代码：
SPFA：

#include <iostream>using namespace std;#include <stdio.h>#include <cstring>#include <queue>#define N 40000#define M 100000#define INF 0x3f3f3f3fstruct node {    int v,w,n;}edge[M];int head[N],dis[N],cnt[N],n,m,s,f,num;int vis[N];void add(int a,int b,int c) {    edge[num].v = b;    edge[num].w = c;    edge[num].n = head[a];    head[a] = num++;}void SPFA() {    for(int i = 0; i < n; i++) {        dis[i] = INF;        vis[i] = 0;    }    dis[s] = 0;    vis[s] = 1;    queue<int> Q;    Q.push(s);    while(!Q.empty()) {        int u = Q.front();        Q.pop();        vis[u] = 0;        for(int i = head[u]; i != -1; i = edge[i].n) {            int v = edge[i].v;            if(dis[v] > dis[u] + edge[i].w) {                dis[v] = dis[u] + edge[i].w;                if(!vis[v]) {                    cnt[v] ++;                    if(cnt[v] > n) {                        f = 1;                        return;                    }                    vis[v] = 1;                    Q.push(v);                }            }        }    }}int main() {    int T;    int u,v,w;    scanf("%d",&T);    while(T--) {        s = 0;    //  t = n - 1;        memset(head,-1,sizeof(head));        memset(cnt,0,sizeof(cnt));        f = 0;        num = 0 ;    //  cout << endl;        scanf("%d%d",&n,&m);        for(int i = 0; i < m; i++) {            scanf("%d%d%d",&u,&v,&w);            add(u,v,w);        }        //cout << endl;        SPFA();        if(f)             printf("possible\n");        else            printf("not possible\n");    }    return 0;}

bellmanford算法：

#include <iostream>using namespace std;#include <cstring>#include <stdio.h>#include <algorithm>const int maxn = 1005;const int maxm = 2005;struct edge {    int u,v;    int d;}e[maxm];const int INF =  0x3f3f3f3f;int N,M;int d[maxn];bool BellmanFord() {    //int flag = 0;    for(int i = 0; i < N; i++) {        d[i] = INF;    }    d[0] = 0;    for(int i = 0; i < N - 1; i++)        for(int j = 0; j < M; j++)            if(d[e[j].u] != INF && d[e[j].v] > d[e[j].u] + e[j].d)                d[e[j].v] = d[e[j].u] + e[j].d;    for(int i = 0; i < M; i++)        if(d[e[i].u]!= INF && d[e[i].u] + e[i].d < d[e[i].v])// {            return true;    /*      flag = 1;            d[e[i].v] = d[e[i].u] + e[i].d;            break;*/        //}    /*if(flag)        return false;    return true;*/    return false;}int main() {    int T;    scanf("%d",&T);    while(T--) {        scanf("%d%d",&N,&M);        for(int i = 0; i < M; i++)            scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].d);        if(BellmanFord())            printf("possible\n");        else            printf("not possible\n");    }    return 0;}

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