Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 11350 | Accepted: 3805 |
Description
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
Output
Sample Input
2
0 0
3 4
3
17 4
19 4
18 50
Sample Output
Scenario #1
Frog Distance = 5.000
Scenario #2
Frog Distance = 1.414
解题思路:prim求最小生成树
算出每个点间的距离,生成邻接矩阵
求出最小生成树中最长的一段距离
一旦目的节点加入最小生成树,结束算法
#include <iostream> #include <cstring> #include <cstdio> #include <cmath> #define MAX_VALUE 1e8 using namespace std; typedef struct { int x; int y; }point; int main() { int times=0; int n; bool visit[201]; int d[201]; point stone[200]; int map[201][201]; int frogDist; while(scanf("%d",&n) && n!=0) { //init times++; memset(visit,0,sizeof(visit)); memset(stone,0,sizeof(stone)); for(int i=1; i<=n; i++) { d[i]=MAX_VALUE; } for(int i=1; i<=n; i++) { scanf("%d%d",&stone[i].x, &stone[i].y); } for(int i=1; i<=n; i++) { for(int j=i+1; j<=n; j++) { map[i][j]=(stone[i].x-stone[j].x)*(stone[i].x-stone[j].x)+(stone[i].y-stone[j].y)*(stone[i].y-stone[j].y); //dist is square of real dist map[j][i]=map[i][j]; } map[i][i]=MAX_VALUE; } frogDist=0; //prim for(int i=1; i<=n; i++) { //Extract min_value node d[1]=0; int minNode=1; int minValue=MAX_VALUE; for(int j=1; j<=n; j++) { if(!visit[j] && d[j]<minValue) { minNode=j; minValue=d[j]; } } if(frogDist<d[minNode]) frogDist=d[minNode]; visit[minNode]=true; if(minNode==2) break; //Finded minNode, and relax in adjcent node for(int j=1; j<=n; j++) { if(!visit[j]&&map[minNode][j]<d[j]) { d[j]=map[minNode][j]; } } } printf("Scenario #%d\n",times); printf("Frog Distance = %.3f\n",sqrt(double(frogDist))); printf("\n"); } return 0; }
附加测试用例:
8
Scenario #1
1 1
4 0
1 2
2 2
3 2
4 2
3 0
5 1
Frog Distance = 1.414
3
Scenario #2
9 10
10 10
100 10
Frog Distance = 1.000