HUD：2853-Assignment（KM算法+hash）

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2853

Assignment

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

Problem Description

Last year a terrible earthquake attacked Sichuan province. About 300,000 PLA soldiers attended the rescue, also ALPCs. Our mission is to solve difficulty problems to optimization the assignment of troops. The assignment is measure by efficiency, which is an integer, and the larger the better.
We have N companies of troops and M missions, M>=N. One company can get only one mission. One mission can be assigned to only one company. If company i takes mission j, we can get efficiency Eij.
We have a assignment plan already, and now we want to change some companies’ missions to make the total efficiency larger. And also we want to change as less companies as possible.

Input

For each test case, the first line contains two numbers N and M. N lines follow. Each contains M integers, representing Eij. The next line contains N integers. The first one represents the mission number that company 1 takes, and so on.
1<=N<=M<=50, 1

Output

For each the case print two integers X and Y. X represents the number of companies whose mission had been changed. Y represents the maximum total efficiency can be increased after changing.

Sample Input

3 3
2 1 3
3 2 4
1 26 2
2 1 3
2 3
1 2 3
1 2 3
1 2

Sample Output

2 26
1 2

---

解题心得：

• 题意就是先给你一个匹配，要你重新匹配，匹配之后的权值最大，并且要求你改变原先匹配次数要最小。

• 匹配出来的权值最大，就是一个KM算法，但是要改变次数最小就要一点小技巧了，原本想的是在原先的匹配方案上加一就可以了，这样在面对大小一样的情况下可以优先选择，但是有个问题就是很可能加一之后和另一个更大值重合了。所以需要先乘以一个很大的值，在加一就可以可以避免重合的问题，也就是hash的问题。然后再直接对比现在的匹配情况和之前的匹配情况。

---

#include<bits/stdc++.h>
using namespace std;
const int maxn = 55;
int maps[maxn][maxn];
int ly[maxn],lx[maxn],match[maxn],pre_match[maxn],slack[maxn];
bool visx[maxn],visy[maxn];
int n,m;

void init()
{
    memset(ly,0,sizeof(ly));
    memset(lx,0,sizeof(lx));
    memset(match,-1,sizeof(match));
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            scanf("%d",&maps[i][j]);
            maps[i][j] *= 100;
            if(maps[i][j] > lx[i])
                lx[i] = maps[i][j];
        }
    for(int i=1;i<=n;i++)
        scanf("%d",&pre_match[i]);
}

bool dfs(int x)
{
    visx[x] = true;
    for(int i=1;i<=m;i++)
    {
        if(visy[i])
            continue;
        int temp;
        temp = lx[x] + ly[i] - maps[x][i];
        if(!temp)
        {
            visy[i] = true;
            if(match[i] == -1 || dfs(match[i]))
            {
                match[i] = x;
                return true;
            }
        }
        else if(temp < slack[i])
            slack[i] = temp;
    }
    return false;
}

int get_pre_sum()
{
    int sum = 0;
    for(int i=1;i<=n;i++)
    {
        sum += maps[i][pre_match[i]]/100;
        maps[i][pre_match[i]]++;
        if(maps[i][pre_match[i]] > lx[i])
            lx[i] = maps[i][pre_match[i]];
    }
    return sum;
}

void KM()
{
    int pre_sum = get_pre_sum();
    for(int i=1;i<=n;i++)
    {
        memset(slack,0x3f,sizeof(slack));
        while(1)
        {
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            if(dfs(i))
                break;

            int temp = 0x3f3f3f3f;
            for(int j=1;j<=m;j++)
                if(!visy[j] && temp > slack[j])
                    temp = slack[j];
            for(int j=1;j<=n;j++)
            {
                if(visx[j])
                    lx[j] -= temp;
            }
            for(int j=1;j<=m;j++)
            {
                if(visy[j])
                    ly[j] += temp;
                else
                    slack[j] -= temp;
            }
        }
    }
    int ans1,ans2;
    ans1 = ans2 = 0;
    for(int i=1;i<=m;i++)
        ans1 += maps[match[i]][i];
    for(int i=1;i<=n;i++)
    {
        if(match[pre_match[i]] != i)//前后对比
            ans2++;
    }
    printf("%d %d
",ans2,ans1/100-pre_sum);
    //也可以写成：printf("%d %d
",ans1%100,ans1/100-pre_sum);
    return ;
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        init();
        KM();
    }
    return 0;
}