Tree
Tree
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1924 Accepted Submission(s): 563
Problem Description
There are N (2<=N<=600) cities,each has a value of happiness,we consider two cities A and B whose value of happiness are VA and VB,if VA is a prime number,or VB is a prime number or (VA+VB) is a prime number,then they can be connected.What's
more,the cost to connecte two cities is Min(Min(VA , VB),|VA-VB|).
Now we want to connecte all the cities together,and make the cost minimal.
Now we want to connecte all the cities together,and make the cost minimal.
Input
The first will contain a integer t,followed by t cases.
Each case begin with a integer N,then N integer Vi(0<=Vi<=1000000).
Each case begin with a integer N,then N integer Vi(0<=Vi<=1000000).
Output
If the all cities can be connected together,output the minimal cost,otherwise output "-1";
Sample Input
2 5 1 2 3 4 5 4 4 4 4 4
Sample Output
4 -1
Author
Teddy
Source
Recommend
#include<stdio.h> #include<string.h> #include<stdlib.h> #define INF 0xfffffff #define min(a,b)(a>b?b:a) int map[1010][1010],p[1000010*2],mark[1010],num[1010]; int n; void fun() { int i,j; p[1]=1; for(i=2;i<1000010*2;i++) { if(!p[i]) { for(j=i+i;j<1000010*2;j+=i) { p[j]=1; } } } } int prim() { int sum=0,p=n,i,j; int flog; memset(mark,0,sizeof(mark)); while(--p) { int min=INF; for(i=2;i<=n;i++) { if(!mark[i]&&map[1][i]<min) { min=map[1][i]; flog=i; } } if(min==INF) break; sum+=min; mark[flog]=1; for(j=2;j<=n;j++) { if(!mark[j]&&map[1][j]>map[flog][j]) map[1][j]=map[flog][j]; } } if(p) return -1; else return sum; } int main() { int t; fun(); scanf("%d",&t); while(t--) { int i,j; scanf("%d",&n); for(i=1;i<=n;i++) scanf("%d",&num[i]); for(i=1;i<=n;i++) for(j=1;j<=n;j++) map[i][j]=INF; for(i=1;i<=n;i++) { for(j=i+1;j<=n;j++) { if(!p[num[i]]||!p[num[j]]||!p[num[i]+num[j]]) { map[j][i]=map[i][j]=min(min(num[i],num[j]),abs(num[i]-num[j])); } } } /*for(i=1;i<=n;i++) for(j=1;j<=n;j++) printf("%d ",map[i][j]);*/ printf("%d\n",prim()); } return 0; }