ZOJ 1942
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Unfortunately Fiona's stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps.
To execute a given sequence of jumps, a frog's jump range obviously must be at least as long as the longest jump occuring in the sequence.
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones.
You are given the coordinates of Freddy's stone, Fiona's stone and all other stones in the lake. Your job is to compute the frog distance between Freddy's and Fiona's stone.
Input
The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2 <= n <= 200). The next n lines each contain two integers xi, yi (0 <= xi, yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy's stone, stone #2 is Fiona's stone, the other n-2 stones are unoccupied. There's a blank line following each test case. Input is terminated by a value of zero (0) for n.
Output
For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.
Sample Input
2
0 0
3 4
3
17 4
19 4
18 5
0
Sample Output
Scenario #1
Frog Distance = 5.000
Scenario #2
Frog Distance = 1.414
#include<stdio.h>#include<math.h>#include<string.h>double dis[1007];double map[1007][1007];int n;void dijkstra(){ int i,j,vis[1007]; memset(vis,0,sizeof(vis)); for(i=1; i<=n; i++) { dis[i]=999999; } dis[1]=0; int pos,now=1; double t; double min; vis[now]=1; for(i=1; i<=n; i++) { min=999999; for(j=1; j<=n; j++) { t=dis[now]>map[now][j]?dis[now]:map[now][j]; if(t<dis[j]) dis[j]=t; } for(j=1; j<=n; j++) if(!vis[j]&&dis[j]<min) { min=dis[j]; now=j; } vis[now]=1; }}int main(){ int textcase=1; while(scanf("%d",&n),n) { int i,j; int a[1005],b[1005]; memset(map,0,sizeof(map)); for(i=1; i<=n; i++) { scanf("%lf%lf",&a[i],&b[i]); } double tmp; for(i=1; i<=n; i++) for(j=i+1; j<=n; j++) { if(i==j) { map[i][j]=999999; continue; } tmp=sqrt(pow((a[j]-a[i]),2.0)+pow((b[j]-b[i]),2.0)); map[i][j]=map[j][i]=tmp; } dijkstra(); printf("Scenario #%d\n",textcase++); printf("Frog Distance = %.3f\n\n",dis[2]); } return 0;}
但是看了看大神的代码,发现用的是结构体然后就AC了,求大神讲解
#include<stdio.h>#include<math.h>#include<string.h>double dis[1007];double map[1007][1007];int n;struct sa{double a,b;}forg[1007];void dijkstra(){ int i,j,vis[1007]; memset(vis,0,sizeof(vis)); for(i=1; i<=n; i++) { dis[i]=999999; } dis[1]=0; int pos,now=1; double t; double min; vis[now]=1; for(i=1; i<=n; i++) { min=999999; for(j=1; j<=n; j++) { t=dis[now]>map[now][j]?dis[now]:map[now][j]; if(t<dis[j]) dis[j]=t; } for(j=1; j<=n; j++) if(!vis[j]&&dis[j]<min) { min=dis[j]; now=j; } vis[now]=1; }}int main(){ int textcase=1; while(scanf("%d",&n),n) { int i,j; //int a[1005],b[1005]; memset(map,0,sizeof(map)); for(i=1; i<=n; i++) { scanf("%lf%lf",&forg[i].a,&forg[i].b); //scanf("%lf%lf",&a[i],&b[i]); } double tmp; for(i=1; i<=n; i++) for(j=i+1; j<=n; j++) { if(i==j) { map[i][j]=999999; continue; } //tmp=sqrt(pow((a[j]-a[i]),2.0)+pow((b[j]-b[i]),2.0)); tmp=sqrt(pow(forg[j].a-forg[i].a,2.0)+pow(forg[j].b-forg[i].b,2.0)); map[i][j]=map[j][i]=tmp; } dijkstra(); printf("Scenario #%d\n",textcase++); printf("Frog Distance = %.3f\n\n",dis[2]); } return 0;}
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