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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;
} 版权声明:本文为博主原创文章,未经博主允许不得转载。
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