HDU 2682 Tree 最小生成树 prim模板

题目：

http://acm.hdu.edu.cn/showproblem.php?pid=2682

题意：

有n个城市，每个城市有一个幸福值，对任意两个城市，如果两个城市中任意一个城市幸福值为素数或者两者之和为素数，那么这两个城市是可以连接的，代价是$min(min(v_i, v_j), |v_i-v_j|)$，求把所有城市连接再一起的最小花费，不能连通输出-1

思路：

最小生成树，测试prim模板
朴素的prim:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;

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

int g[N][N];
int mincost[N];
bool vis[N], prime[M];

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], g[v][i]);
    }
    return num == n ? ans : -1;
}
void table()
{
    prime[0] = prime[1] = 1;
    for(int i = 2; i * i < M; i++)
        if(! prime[i])
            for(int j = i + i; j < M; j += i) prime[j] = 1;
}
int main()
{
    table();
    int t, n, a[N];
    scanf("%d", &t);
    while(t--)
    {
        memset(g, 0x3f, sizeof g);
        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]));
                    g[i][j] = g[j][i] = cost;
                }
            }
        printf("%d\n", prim(n));
    }
    return 0;
}

二叉堆优化的prim:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;

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

struct edge
{
    int to, cost, next;
}g[N*N*2];

int cnt, head[N];
int mincost[N];
bool vis[N], prime[M];

void add_edge(int v, int u, int cost)
{
    g[cnt].to = u, g[cnt].cost = cost, g[cnt].next = head[v], head[v] = cnt++;
}
int prim(int n)
{
    priority_queue<pii, vector<pii>, greater<pii> > que;
    memset(vis, 0, sizeof vis);
    memset(mincost, 0x3f, sizeof mincost);
    que.push(pii(0, 1)), mincost[1] = 0;
    int ans = 0, num = 0;
    while(! que.empty())
    {
        int v = que.top().second; que.pop();
        if(vis[v]) continue;
        vis[v] = true;
        num++;
        ans += mincost[v];
        for(int i = head[v]; i != -1; i = g[i].next)
        {
            int u = g[i].to;
            if(mincost[u] > g[i].cost)
            {
                mincost[u] = g[i].cost;
                que.push(pii(mincost[u], u));
            }
        }
    }
    return num == n ? ans : -1;
}
void table()
{
    prime[0] = prime[1] = 1;
    for(int i = 2; i * i < M; i++)
        if(! prime[i])
            for(int j = i + i; j < M; j += i) prime[j] = 1;
}
int main()
{
    table();
    int t, n, a[N];
    scanf("%d", &t);
    while(t--)
    {
        cnt = 0;
        memset(head, -1, sizeof head);
        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]));
                    add_edge(i, j, cost), add_edge(j, i, cost);
                }
            }
        printf("%d\n", prim(n));
    }
    return 0;
}