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Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9377 Accepted Submission(s): 3313
3 1 0.5 0.5 0.5 1 0.4 0.5 0.4 1 3 1 2 2 3 1 3
0.500 0.400 0.500
#include <stdio.h> #include <string.h> #include <queue> #include <algorithm> #include <iostream> #define INF 0x3f3f3f3f using namespace std; int n; int edgenum; int head[1000005]; double dis[1005]; int vis[1005]; struct Edge { int from,to; double val; int next; }edge[1000005]; void init() { edgenum=0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,double w) { Edge E={u,v,w,head[u]}; edge[edgenum]=E; head[u]=edgenum++; } void getmap() { double a; for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { scanf("%lf",&a); addedge(i,j,a); } } } void SPFA(int x) { queue<int>q; q.push(x); memset(vis,0,sizeof(vis)); memset(dis,0,sizeof(dis));//这一点要全部赋值为0,因为刚开始假设所有的点都不能到 dis[x]=1;//就只能到达起点! vis[x]=1; while(!q.empty()) { int u=q.front(); q.pop(); vis[u]=0;//重新赋值0! for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].to; if(dis[v]<dis[u]*edge[i].val)//要选择安全系数大的点,所以要将大的值赋值给dis数组! { dis[v]=dis[u]*edge[i].val; if(vis[v]==0) { vis[v]=1; q.push(v);//进而用大的值去更新其他的值! } } } } } int main() { while(scanf("%d",&n)!=EOF) { init(); getmap(); int m,x,y; scanf("%d",&m); for(int i=1;i<=m;i++) { scanf("%d%d",&x,&y); SPFA(x); if(dis[y]==0) printf("What a pity!\n"); else printf("%.3lf\n",dis[y]); } } return 0; }
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HDU 1596 find the safest road <SPFA算法的一个变形>
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原文地址:http://blog.csdn.net/dxx_111/article/details/47804303