HDU 1875 畅通工程再续

 

http://acm.hdu.edu.cn/showproblem.php?pid=1875

 

解题思路：还是用到prim算法啦，只不过这道题他没给你两点之间的权值，需要自己用for循环一个个求出i点到j点的距离，时间复杂度就是O（n^2）了，之后就是标准的prim算法做啦。

 

#include <stdio.h>#include <string.h>#include <math.h>#define INIT 999999999#define MaxVertex 105int NumOfVertex;int count;double sum;double Graph[MaxVertex][MaxVertex];struct Point{int x;int y;}points[MaxVertex];void Prim(){double distance[MaxVertex];bool visited[MaxVertex];sum = 0;memset(visited,0,sizeof(visited));int i,j;double min;int locate;for (i=0;i<NumOfVertex;i++)distance[i] = Graph[0][i];visited[0] = true;count = 1;for (j=0;j<NumOfVertex;j++){min = INIT;for (i=1;i<NumOfVertex;i++){if (!visited[i]&&distance[i]<min){min = distance[i];locate = i;}}if(min!=INIT){count++;visited[locate] = true;sum+=min;for (i=0;i<NumOfVertex;i++){if (!visited[i]&&distance[i]>Graph[locate][i]){distance[i] = Graph[locate][i];}}}}}double dist(Point a,Point b){double k;k = sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));if (k<=1000&&k>=10)return k;return INIT;}int main(){int i,j;int NumOfTest;/*freopen("e://1.txt","r",stdin);*/while (scanf("%d",&NumOfTest)!=EOF){while (NumOfTest--){scanf("%d",&NumOfVertex);for (i=0;i<NumOfVertex;i++){for(j=0;j<NumOfVertex;j++)if(i==j)Graph[i][j] = 0;elseGraph[i][j] = INIT;}for(i=0;i<NumOfVertex;i++)scanf("%d%d",&points[i].x,&points[i].y);for(i=0;i<NumOfVertex;i++){for(j=0;j<NumOfVertex;j++)if(i!=j) Graph[i][j] = 100*(dist(points[i],points[j]));}Prim();if (count == NumOfVertex){printf("%0.1lf/n",sum);}elseprintf("oh!/n");}}return 0;}