Assignment HDU

Assignment

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1698    Accepted Submission(s): 901

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<Eij<=10000.
Your program should process to the end of file.

 

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

 

Source

2009 Multi-University Training Contest 5 - Host by NUDT

 

Recommend

gaojie

 

关键是求最小的改变 （尽量减少公司改变任务 且 使效率最大）

做法就和求最小割边的条数一样

我们把每个w都乘上100（大于n即可 一会再解释）

然后把对每个公司原分配的w + 1

那么在同等条件下 原分配的任务是不是  就会优先被选择

而乘100

最后求出max_value % 100 是不是就是原分配的任务 在最后分配中保留的个数  （因为除了原分配的任务 都是100的倍数啊）

那么max_value / 100 是不是就是最后分配效率的真实值 

有人可能会问有的值不是+ 1 了吗

就算n个数 全 + 1， 那是不是 + n  ， n 除一个大于n的数是不是就是0

这就是为什么要乘一个比n大的数的原因

然后就是裸km了

 

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d
", a)
#define plld(a) printf("%lld
", a)
#define pc(a) printf("%c
", a)
#define ps(a) printf("%s
", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 55, INF = 0x7fffffff;
int n, m;
int usedx[maxn], usedy[maxn], w[maxn][maxn], bx[maxn], by[maxn], cx[maxn], cy[maxn], slack[maxn];
int nx, ny, minn, max_value;

int dfs(int u)
{
    usedx[u] = 1;
    for(int i=1; i<=ny; i++)
    {
        if(usedy[i] == -1)
        {
            int t = bx[u] + by[i] - w[u][i];
            if(t == 0)
            {
                usedy[i] = 1;
                if(cy[i] == -1 || dfs(cy[i]))
                {
                    cy[i] = u;
                    cx[u] = i;
                    return 1;
                }
            }
            else
                slack[i] = min(slack[i], t);
        }
    }
    return 0;
}

int km()
{
    mem(cx, -1);
    mem(cy, -1);
    mem(bx, -1);
    mem(by, 0);
    for(int i=1; i<=nx; i++)
        for(int j=1; j<=ny; j++)
            bx[i] = max(bx[i], w[i][j]);
    for(int i=1; i<=nx; i++)
    {
        for(int j=1; j<=ny; j++)
            slack[j] = INF;
        while(1)
        {
            mem(usedx, -1);
            mem(usedy, -1);
            if(dfs(i)) break;
            int d = INF;
            for(int j=1; j<=ny; j++)
                if(usedy[j] == -1)
                    d = min(d, slack[j]);
            for(int j=1; j<=nx; j++)
                if(usedx[j] != -1) bx[j] -= d;
            for(int j=1; j<=ny; j++)
                if(usedy[j] != -1) by[j] += d;
                else
                    slack[j] -= d;
        }
    }
    max_value = 0;
    for(int i=1; i<=nx; i++)
        if(cx[i] != -1)
            max_value += w[i][cx[i]];
    return max_value;
}


int main()
{
    while(scanf("%d%d", &n, &m) != EOF)
    {
        nx = n, ny = m;
        int tmp;
        rap(i, 1, n)
            rap(j, 1, m)
            {
                rd(w[i][j]);
                w[i][j] *= 100;
            }
        int pre_sum = 0;
        rap(i, 1, n)
        {
            rd(tmp);
            pre_sum += w[i][tmp];
            w[i][tmp] += 1;
        }
        max_value = km();
        printf("%d %d
", n - max_value % 100, max_value / 100 - pre_sum / 100);

    }

    return 0;
}

Assignment

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1698    Accepted Submission(s): 901

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<Eij<=10000.
Your program should process to the end of file.

 

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

 

Source

2009 Multi-University Training Contest 5 - Host by NUDT

 

Recommend

gaojie