HDU 2682
2014-04-30 18:35
411 查看
思路:由于题目对能相连的点有限制,必须将这些点处理,能相连的点合并到一个集合中,最后查看是否所有点都在一个集合里,若都在说明是一个连通图,存在最小生成树,否则图不连通,不存在最小花费。
#include<stdio.h> #include<string.h> #include<stdlib.h> int CityHappy[605],vis[605]; int isprime[1000005],dist[605]; int map[601][601],n; int father[605],depth[605]; void init_B() { int i; for(i = 1;i <= n;i ++) { father[i] = i; depth[i] = 0; } } int find(int x) { if(x == father[x]) return x; return father[x] = find(father[x]); } void unit(int x,int y) { x = find(x); y = find(y); if(x == y) return ; if(depth[x] > depth[y]) father[y] = x; else { if(depth[x] < depth[y]) father[x] = y; else { father[x] = y; depth[y]++; } } } void prime() { int i,j; isprime[0] = isprime[1] = 1; for(i = 2;i <= 1e6;i ++) { if(!isprime[i]) { for(j = i << 1;j <= 1e6;j += i) isprime[j] = 1; } } } int judge(int a,int b) { if(!isprime[a] || !isprime[b]) return 1; if(!isprime[a+b]) return 1; return 0; } int min(int a,int b) { return a < b?a:b; } void opration() { int i,j,a,b; init_B(); for(i = 1;i <= n;i ++) { for(j = 1;j <= n;j ++) { if(i != j) { a = CityHappy[i]; b = CityHappy[j]; if(judge(a,b)) { map[i][j] = min(min(a,b),abs(a-b)); map[j][i] = map[i][j]; unit(i,j); } else map[i][j] = map[j][i] = 1 << 30; } } } } void init() { int i; memset(vis,0,sizeof(vis)); for(i = 1;i <= n;i ++) dist[i] = map[1][i]; } int main() { int t,i,j,k,cnt,min,sum; scanf("%d",&t); prime(); while(t--) { sum = cnt = 0; scanf("%d",&n); for(i = 1;i <= n;i ++) scanf("%d",&CityHappy[i]); opration(); init(); for(i = 1;i <= n;i ++) { if(i == find(i)) cnt++; if(cnt == 2) break; } if(cnt == 2) { printf("-1\n"); continue ; } for(i = 0;i < n;i ++) { min = 1 << 30; for(j = 1;j <= n;j ++) { if(!vis[j] && min > dist[j]) { min = dist[j]; k = j; } } vis[k] = 1; if(min != 1 << 30) sum += min; for(j = 1;j <= n;j ++) { if(!vis[j] && dist[j] > map[k][j]) dist[j] = map[k][j]; } } printf("%d\n",sum); } return 0; }