Connect the Cities（prim）用prim都可能超时，交了20几发卡时过的

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 701 Accepted Submission(s): 212

Problem Description

In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.  

 

Input

The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.

 

Output

For each case, output the least money you need to take, if it’s impossible, just output -1.

 

Sample Input

1
6 4 3
1 4 2
2 6 1
2 3 5
3 4 33
2 1 2
2 1 3
3 4 5 6

 

Sample Output

1

 

Author

dandelion

 

Source

HDOJ Monthly Contest – 2010.04.04

 

Recommend

lcy

/*
点多变少的时候用克利斯卡尔算法，点少边的时候用prim算法

真心的心累，交了20几发卡时过的
*/
#include<bits/stdc++.h>
#define N 550
#define INF 0x3f3f3f3f
using namespace std;
/*********输入输出外挂************/
template <class T>
inline bool scan_d(T &ret)
{
    char c;
    int sgn;
    if(c=getchar(),c==EOF)
        return 0; //EOF
    while(c!='-'&&(c<'0'||c>'9'))
        c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9')
        ret=ret*10+(c-'0');
    ret*=sgn;
    return 1;
}
inline void out(int x)
{
    if(x>9)
        out(x/10);
    putchar(x%10+'0');
}
/*********输入输出外挂************/
int t;
int n,m,k;
int mapn[550][550];//用来构建图
bool vis[550];//用来记录那个点访问过也就是判环用的
int d[550];//表示选中的结点到当前结点的最小距离
int x,y,val,q,rt[550];
void init()
{
    memset(mapn,INF,sizeof mapn);
    memset(vis,false,sizeof vis);
    memset(d,INF,sizeof d);
}
int prim(int m)
{
    int root;//表示当前选中的结点
    int minn=INF;
    root=m;
    vis[m]=true;
    long long cur=0;
    int len=0;
    for(int i=2;i<=n;i++)//优化的prim算法，总共要找n个点嘛，先找了m了，然后再找n-1个点就够了
    {
        minn=INF;
        for(int j=1;j<=n;j++)//然后开始找点
            if(!vis[j])//这个点没有遍历过
            {
                if(d[j]>mapn[root][j])
                    d[j]=mapn[root][j];
                if(minn>d[j])
                {
                    minn=d[j]; m=j;
                }
            }
        if(!vis[m])//这个点没有遍历过
        {
            cur+=minn;
            vis[m]=true;
            root=m;
            len++;

        }
    }
    if(len==n-1)
        return cur;
    return -1;
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    scan_d(t);
    while(t--)
    {
        scan_d(n);scan_d(m);scan_d(k);
        init();//初始化
        for(int i=0;i<m;i++)
        {
            scan_d(x);scan_d(y);scan_d(val);
            if(mapn[x][y]>val)//判重边
            {
                mapn[x][y]=mapn[y][x]=val;
            }
        }
        for(int i=0;i<k;i++)
        {
            scan_d(q);
            for(int j=0;j<q;j++)
                scan_d(rt[j]);
            /*既然已经联通了那么相互之间就不需要再花费钱去修路了*/
            for(int j=0;j+1<q;j++)
                    mapn[rt[j]][rt[j+1]]=mapn[rt[j+1]][rt[j]]=0;
        }
        printf("%d\n",prim(1));
    }
    return 0;
}