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http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?cid=963&pid=1019&ojid=1

2011-11-04 13:19 525 查看
这一题是dijstra的变种,,,用的是dijstra的思想和方法,让求的是从一点到另一点的最大的最小值,,,,一开始木有认真读题,,,贡献了5次wa,,,
#include <iostream>
#include <cmath>
#include<cstdio>
using namespace std;
#define MAX 201
#define INF 10000.0f
float matrix[MAX][MAX];
int s[MAX];
float dis[MAX];
typedef struct point
{
int x;
int y;
}point;
void Dijkstra(int &n)
{         int now=1;
for(int i = 1; i <= n; ++i)
{
dis[i] = matrix[now][i];
s[i] = 0;
}
s[now] = 1;
for(int i = 1; i <= n-1; ++i)
{
float minDis = INF;
for(int j = 1; j <= n; ++j)
if(!s[j] && dis[j] < minDis)
minDis = dis[now=j];
s[now] = 1;
for(int j = 1; j <= n; ++j)
if(!s[j] && matrix[now][j]!= INF)
dis[j] = min(dis[j],max(dis[now],matrix[now][j]));
}
}
int main()
{
int n;
point p[MAX];
int count = 0;
while(~scanf("%d",&n)&& n)
{for(int i = 1; i <= n; ++i)
~scanf("%d%d", &p[i].x, &p[i].y);
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
if(i!=j)  matrix[i][j] = INF;
else    matrix[i][j] = 0;
for(int i = 1; i < n; ++i)
for(int j = i+1; j <=n; ++j)
matrix[j][i] = matrix[i][j] = sqrt(((float)(p[i].x-p[j].x)*(p[i].x-p[j].x))
+ (p[i].y-p[j].y)*(p[i].y-p[j].y));

Dijkstra(n);
printf("Scenario #%d\nFrog Distance = %.3f\n\n", ++count, dis[2]);
}
return 0;
}
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标签:  matrix distance float struct