HDOJ 2682 Tree

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（有米！）

Tree

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2037 Accepted Submission(s): 590

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.

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).

Output

If the all cities can be connected together,output the minimal cost,otherwise output "-1";

Sample Input

251234544444

Sample Output

4-1

Author

Teddy

最小生成树，用素数筛法打表，判断素数就可以了

#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;}

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