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英语太渣读了半天,理解了题意就好办了……Bellman_ford算法。在n-1次松弛后,如果依然存在满足松弛的情况返回1.说简单点就是判读是否存在正环。
#include<stdio.h> #include<string.h> #include<cstring> #include<string> #include<math.h> #include<queue> #include<algorithm> #include<iostream> #include<stdlib.h> #include<cmath> #define INF 0x3f3f3f3f #define MAX 1000005 using namespace std; struct node { int a,b; double e,c; }Map[MAX]; int n,m,s,k; double dist[MAX],v; int bellman_ford() { int i,j,ok; memset(dist,0,sizeof(dist)); dist[s]=v; for(i=1;i<n;i++) { ok=0; for(j=1;j<k;j++) { if(dist[Map[j].b] < (dist[Map[j].a] - Map[j].c)*Map[j].e) { dist[Map[j].b] = (dist[Map[j].a] - Map[j].c)*Map[j].e; ok=1; } } if(!ok) break; } for(j=1;j<k;j++) { if(dist[Map[j].b] < (dist[Map[j].a] - Map[j].c)*Map[j].e) return 1; } return 0; } int main() { int i,j,a,b; double e1,c1,e2,c2; while(scanf("%d%d%d%lf",&n,&m,&s,&v)!=EOF) { k=1; for(i=1;i<=m;i++) { scanf("%d%d%lf%lf%lf%lf",&a,&b,&e1,&c1,&e2,&c2); Map[k].a=a; Map[k].b=b; Map[k].e=e1; Map[k++].c=c1; Map[k].a=b; Map[k].b=a; Map[k].e=e2; Map[k++].c=c2; } int ok=bellman_ford(); if(ok) printf("YES\n"); else printf("NO\n"); } return 0; }
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原文地址:http://www.cnblogs.com/alan-W/p/5665260.html