poj 3259 uva 558 Wormholes(bellman最短路负权回路判断)

poj 3259：

题意：John的农场里n块地，m条路连接两块地，w个虫洞，虫洞是一条单向路，不但会把你传送到目的地，而且时间会倒退Ts。

任务是求你会不会在从某块地出发后又回来，看到了离开之前的自己。

判断树中是否存在负权回路就ok了。

bellman代码：

#include<stdio.h>const int MaxN = 501;//农场数const int INF = 0x3f3f3f3f;int n;//点个数int l;//边标记struct edge{    int from;    int to;    int len;}e[5201];//2500条路径，2500*2条边+200个虫洞bool Bellman(){    int dis[MaxN];    for(int i = 1; i <= n; i++)        dis[i] = INF;    dis[1] = 0;    for(int i = 1; i < n; i++)    {        bool over = true;//判断是否松弛完毕        for(int j = 0; j < l; j++)        {            int from = e[j].from;            int to = e[j].to;            int len = e[j].len;            if(dis[from] != INF && dis[to] > dis[from] + len)            {                dis[to] = dis[from] + len;                over = false;            }        }        if(over)            break;    }    for(int j = 0; j < l; j++)    {        int from = e[j].from;        int to = e[j].to;        int len = e[j].len;        if(dis[to] > dis[from] + len)            return true;    }    return false;}int main(){    //freopen("in.txt", "r", stdin);    int ncase;    int m, w;    scanf("%d", &ncase);    while(ncase--)    {        l = 0;        scanf("%d%d%d", &n, &m, &w);        int from, to, len;        for(int i = 0; i < m; i++)        {            scanf("%d%d%d", &from, &to, &len);            e[l].from = from;            e[l].to = to;            e[l++].len = len;            e[l].from = to;            e[l].to = from;            e[l++].len = len;        }        for(int i = 0; i < w; i++)        {            scanf("%d%d%d", &from, &to, &len);            e[l].from = from;            e[l].to = to;            e[l++].len = -len;        }        if(Bellman())            printf("YES\n");        else            printf("NO\n");    }    return 0;}

uva 558：

差不多的题目。

代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <cmath>#include <stack>#include <vector>#include <queue>#include <map>#include <climits>#include <cassert>#define LL long longusing namespace std;const int maxn = 1000 + 10;const int maxm = 2000 + 10;const int inf = 0x3f3f3f3f;struct Edge{    int fr, to, cost;} e[maxm];int n, m;bool bellman(){    int dis[maxn];    memset(dis, inf, sizeof(dis));    dis[0] = 0;    for (int i = 0; i < n - 1; i++)    {        bool over = true;        for (int j = 0; j < m; j++)        {            int fr = e[j].fr;            int to = e[j].to;            int cost = e[j].cost;            if (dis[fr] != inf && dis[fr] + cost < dis[to])            {                dis[to] = dis[fr] + cost;                over = false;            }        }        if (over)            break;    }    for (int j = 0; j < m; j++)    {        int fr = e[j].fr;        int to = e[j].to;        int cost = e[j].cost;        if (dis[fr] + cost < dis[to])            return true;    }    return false;}int main(){    #ifdef LOCAL    freopen("in.txt", "r", stdin);    #endif // LOCAL    int ncase;    scanf("%d", &ncase);    while (ncase--)    {        scanf("%d%d", &n, &m);        for (int i = 0; i < m; i++)        {            int u, v, w;            scanf("%d%d%d", &u, &v, &w);            e[i].fr = u;            e[i].to = v;            e[i].cost = w;//            e[i + m].fr = v;//            e[i + m].to = u;//            e[i + m].cost = w;        }        if (bellman())            printf("possible\n");        else            printf("not possible\n");    }    return 0;}