Bellman 算法

标签：最短路   bellman   

这道题目其实就是在求有没有正环。与求负环的区别就是要不断的更新值，但是这个值要变大。而不是变小。

Description

Input

Output

Sample Input

3 2 1 20.0
1 2 1.00 1.00 1.00 1.00
2 3 1.10 1.00 1.10 1.00

Sample Output

YES

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <set>
#include <map>
using namespace std;
const int INF = 99999999.9;
const int EM = 5555;
const int VM = 110;
int n,m,s;
double mon;

struct Edge{
    int u,v;
    double com;
    double exg;
}edge[EM<<1];
int cnt;
double dis[VM];
//double mon;

/*void addage(int cu,int vu,double aa,double bb){
    edge[cnt].from = cu;
    edge[cnt].to = vu;
    edge[cnt].com = aa;
    edge[cnt].exg = bb;
    cnt++;
}*/

int Bellman(){
    int i,j,flag;
    for(i=1;i<=n;i++){
        dis[i] = 0.0;
    }
    dis[s] = mon;
    for(i=1;i<n;i++){
        bool flag = false;
        for(j=0;j<cnt;j++){
            int u = edge[j].u;
            int v = edge[j].v;
            double rate = edge[j].exg;
            double cost = edge[j].com;
            if(dis[v]<(dis[u]-cost)*rate)//求最大的路径
            {
                dis[v]=(dis[u]-cost)*rate;
                flag=true;
            }
            //if(dis[edge[j].to] < (dis[edge[j].from]-edge[j].com)*edge[j].exg){
            //    dis[edge[j].to] = (dis[edge[j].from]-edge[j].com)*edge[j].exg;
            //    flag = 1;
            //}
        }
        if(flag==false){
            return false;
        }
    }
    for(j=0; j<cnt; j++)
    {
        if(dis[edge[j].v]<(dis[edge[j].u]-edge[j].com)*edge[j].exg)//与传统的bell不一样，传统的bell是找负环。现在是找正环，正环无限松弛
            return true;
    }
    return false;
}

int main(){
    //int n,m,s;
    //double mon;
    double rab,cab,rba,cba;
    int marka,markb;

    while(~scanf("%d%d%d%lf",&n,&m,&s,&mon)){
        cnt = 0;
        //flag = 1;
        while(m--){
            scanf("%d%d%lf%lf%lf%lf",&marka,&markb,&rab,&cab,&rba,&cba);
            //addage(marka,markb,a2,a1);
            //addage(markb,marka,b2,b1);
            edge[cnt].u=marka,edge[cnt].v=markb,edge[cnt].com=cab,edge[cnt].exg=rab;
            cnt++;
            edge[cnt].u=markb,edge[cnt].v=marka,edge[cnt].com=cba,edge[cnt].exg=rba;
            cnt++;
        }
        if(Bellman()){
            printf("YES\n");
        }
        else{
            printf("NO\n");
        }
    }

    return 0;
}

Bellman 算法

标签：最短路   bellman   

原文地址：http://blog.csdn.net/zcr_7/article/details/39777965