[HDU

最小生成树。

题意就是有N座城市，每个城市有一定的幸福值a[i]。对于任意两个城市i和j，如果a[i]，a[j]，a[i]+a[j]中任意一者的值为素数，那么他们的边权就是min(min(a[i],a[j]),abs(a[i]-a[j]))。问题就是这一幅图的最小生成树。
显然，边一旦建出来了，这就是一道裸题。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

typedef pair<int, int> pii;
const int N = 600 + 10, maxn = 2000000 + 10, inf = 0x3f3f3f3f;

int graph[N][N];
int mincost[N];
int vis[N];
int prime[maxn];

int prim(int n)
{
    memset(mincost, 0x3f, sizeof mincost);
    memset(vis, 0, sizeof vis);
    mincost[1] = 0;
    int ans = 0, num = 0;
    while(true)
    {
        int v = -1;
        for(int i = 1; i <= n; i++)
            if(! vis[i] && (v == -1 || mincost[i] < mincost[v])) v = i;
        if(v == -1 || mincost[v] == inf) 
            break;
        vis[v] = true;
        ans += mincost[v];
        num++;
        for(int i = 1; i <= n; i++) 
            mincost[i] = min(mincost[i], graph[v][i]);
    }
    return num == n ? ans : -1;
}
void filter()
{
    prime[0] = prime[1] = 1;
    for(int i = 2; i * i < maxn; i++)
        if(! prime[i])
            for(int j = i + i; j < maxn; j += i) 
                prime[j] = 1;
}
int main()
{
    filter();
    int t, n, a[N];
    scanf("%d", &t);
    while(t--)
    {
        memset(graph, 0x3f3f3f, sizeof(graph));
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) 
            scanf("%d", &a[i]);
        for(int i = 1; i <= n; i++)
            for(int j = i + 1; j <= n; j++)
            {
                if(! prime[a[i]] || ! prime[a[j]] || ! prime[a[i]+a[j]])
                {
                    int cost = min(min(a[i], a[j]), abs(a[i] - a[j]));
                    graph[i][j] = graph[j][i] = cost;
                }
            }
        printf("%d
", prim(n));
    }
    return 0;
}