# 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

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gaojie

#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\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", 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", 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

Recommend
gaojie