灌水(最小生成树)(Irrigation (minimum spanning tree))

Farmer John已经决定把水灌到他的n(1<=n<=300)块农田,农田被数字1到n标记。把一块土地进行灌水有两种方法,从其他农田饮水,或者这块土地建造水库。建造一个水库需要花费wi(1<=wi<=100000),连接两块土地需要花费Pij(1<=pij<=100000,pij=pji,pii=0). 计算Farmer John所需的最少代价。

Input

*第一行:一个数n

*第二行到第n+1行:第i+1行含有一个数wi

*第n+2行到第2n+1行:第n+1+i行有n个被空格分开的数,第j个数代表pij。

Output

*第一行:一个单独的数代表最小代价.

Sample Input

4

5

4

4

3

0 2 2 2

2 0 3 3

2 3 0 4

2 3 4 0

Sample Output

9

输出详解:

Farmer John在第四块土地上建立水库,然后把其他的都连向那一个,这样就要花费3+2+2+2=9

解析:

本题既有边权,又有点权,如何将它们统一起来呢?不妨设置一个超级水源点,将它与各个点连边,代价为在那个点修水库所需的代价。接下来就可以使用最小生成树算法进行计算最小代价了。

代码如下:

#include<bits/stdc++.h>
using namespace std;
int n,m1,m2,aa;
int v[310][310],lowest[310],z=0;
bool d[310];
void mst()
{
    int ans=0;
    memset(d,true,sizeof(d));
    for(int i=1;i<=n;i++)
        lowest[i]=2147483647;
    lowest[1]=0;
    while(true)
    {
        m1=2147483647;
        m2=0;
        for(int i=1;i<=n;i++)
            if(d[i]==true&&lowest[i]<m1)
            {
                m2=i;
                m1=lowest[i];
            }
        ans=ans+m1;
        d[m2]=false;
        z++;
        if(z==n)
            break;
        for(int i=1;i<=n;i++)
            if(d[i]==true&&v[m2][i]<lowest[i])
                lowest[i]=v[m2][i]; 
 
    }
    cout<<ans<<endl;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)//构造一个n+1点,从它引边到每一个点,边权值为建每一个点的代价
    {
        scanf("%d",&v[i][n+1]);
        v[n+1][i]=v[i][n+1];
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            scanf("%d",&v[i][j]);
    n++;        
    mst();      
    //cout<<ans;
    return 0;
}
————————

Farmer John has decided to pour water into his n (1 & lt; = n & lt; = 300) farmland, which is marked by the numbers 1 to n. There are two ways to irrigate a piece of land, drinking water from other farmland, or building a reservoir on the land. It costs wi (1 & lt; = wi & lt; = 100000) to build a reservoir and PIJ (1 & lt; = PIJ & lt; = 100000, PIJ = PJI, PII = 0) to connect two lands Calculate the minimum cost of Farmer John.

Input

*First line: a number n

*Line 2 to line n + 1: line I + 1 contains a number wi

*Line n + 2 to line 2n + 1: line n + 1 + I has n numbers separated by spaces, and the j number represents PIJ.

Output

*Line 1: a single number represents the minimum cost

Sample Input

four

five

four

four

three

0 2 2 2

2 0 3 3

2 3 0 4

2 3 4 0

Sample Output

nine

Output details:

Farmer John builds a reservoir on the fourth land, and then connects the others to that one. In this way, it costs 3 + 2 + 2 + 2 = 9

Resolution:

This topic has both edge power and point power. How to unify them? You might as well set up a super water source point and connect it with each point at the cost of building a reservoir at that point. Next, the minimum spanning tree algorithm can be used to calculate the minimum cost.

The code is as follows:

#include<bits/stdc++.h>
using namespace std;
int n,m1,m2,aa;
int v[310][310],lowest[310],z=0;
bool d[310];
void mst()
{
    int ans=0;
    memset(d,true,sizeof(d));
    for(int i=1;i<=n;i++)
        lowest[i]=2147483647;
    lowest[1]=0;
    while(true)
    {
        m1=2147483647;
        m2=0;
        for(int i=1;i<=n;i++)
            if(d[i]==true&&lowest[i]<m1)
            {
                m2=i;
                m1=lowest[i];
            }
        ans=ans+m1;
        d[m2]=false;
        z++;
        if(z==n)
            break;
        for(int i=1;i<=n;i++)
            if(d[i]==true&&v[m2][i]<lowest[i])
                lowest[i]=v[m2][i]; 
 
    }
    cout<<ans<<endl;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)//构造一个n+1点,从它引边到每一个点,边权值为建每一个点的代价
    {
        scanf("%d",&v[i][n+1]);
        v[n+1][i]=v[i][n+1];
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            scanf("%d",&v[i][j]);
    n++;        
    mst();      
    //cout<<ans;
    return 0;
}