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hdu 2255 奔小康赚大钱【最大权匹配】

时间:2015-06-15 22:17:03      阅读:123      评论:0      收藏:0      [点我收藏+]

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题目链接:http://acm.acmcoder.com/showproblem.php?pid=2255
题意:中文
//KM算法模板题,用来测试一下模板
代码:

#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>

using namespace std;

const int N = 310;
const int INF = 0x3f3f3f3f;

int nx, ny;

int g[N][N];
int linker[N], lx[N], ly[N];
int slack[N];
bool visx[N], visy[N];

bool dfs(int x)
{
    visx[x] = true;
    for (int y = 0; y < ny; y++)
    {
        if (visy[y]) continue;
        int tmp = lx[x] + ly[y] - g[x][y];
        if (tmp == 0)
        {
            visy[y] = true;
            if (linker[y] == -1 || dfs(linker[y]))
            {
                linker[y] = x;
                return true;
            }
        }
        else if (slack[y] > tmp)
            slack[y] = tmp;
    }
    return false;
}

int KM()
{
    memset(linker,-1,sizeof(linker));
    memset(ly,0,sizeof(ly));
    for (int i = 0; i < nx; i++)
    {
        lx[i] = -INF;
        for (int j = 0; j < ny; j++)
        {
            if (g[i][j] > lx[i])
                lx[i] = g[i][j];
        }
    }
    for (int x = 0; x < nx; x++)
    {
        for (int i = 0; i < ny; i++)
            slack[i] = INF;
        while (true)
        {
            memset(visx, false, sizeof(visx));
            memset(visy, false, sizeof(visy));
            if (dfs(x)) break;
            int d = INF;
            for (int i = 0; i < ny; i++)
                if (!visy[i] && d > slack[i])
                    d = slack[i];
            for (int i = 0; i < nx; i++)
                if (visx[i])
                    lx[i] -= d;
            for (int i = 0; i < ny; i++)
            {
                if (visy[i]) ly[i] += d;
                else slack[i] -= d;
            }
        }
    }
    int res = 0;
    for (int i = 0; i < ny; i++)
    {
        if (linker[i] != -1)
            res += g[linker[i]][i];
    }
    return res;
}

int main()
{
    int n;
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                scanf("%d",&g[i][j]);
        nx = ny = n;
        printf("%d\n",KM());
    }
    return 0;
}

hdu 2255 奔小康赚大钱【最大权匹配】

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原文地址:http://blog.csdn.net/u014427196/article/details/46506525

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