HDU 2682

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;}