POJ

题目链接

http://poj.org/problem?id=2195

题意
在一张N * M 的地图上 有 K 个人 和 K 个房子
地图上每个点都是认为可行走的 求 将每个人都分配到不同的房子 求他们的总的最小步数

思路
因为每个点都是可行走的 我们可以直接根据坐标 算出 每个人都不同房子的路径 然后用 KM 算法跑一下就可以了

KM算法 参考

https://blog.csdn.net/thundermrbird/article/details/52231639
https://blog.csdn.net/pi9nc/article/details/12250247

AC代码

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<list>
#include<stack>
#include <queue>

#define CLR(a, b) memset(a, (b), sizeof(a))

using namespace std;
typedef long long ll;
typedef pair <int, int> pii;
const int INF = 0x3f3f3f3f;
const int maxn = 1e2 + 5;
const int MOD = 1e9;

int nx, ny;
int linker[maxn], lx[maxn], ly[maxn];
int g[maxn][maxn];
int slack[maxn];
bool visx[maxn], visy[maxn];

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()
{
    CLR(linker, -1);
    CLR(ly, 0);
    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)
        {
            CLR(visx, false);
            CLR(visy, false);
            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 dis(int x1, int y1, int x2, int y2)
{
    int x = abs(x1 - x2);
    int y = abs(y1 - y2);
    return x + y;
}


int main()
{
    int n, m;
    while (scanf("%d%d", &n, &m) && (n || m))
    {
        vector <pii> p, h;  
        p.clear();
        h.clear();
        char c;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                scanf(" %c", &c);
                if (c == 'm')
                    p.push_back (pii(i, j));
                if (c == 'H')
                    h.push_back (pii(i, j));
            }
        }
        int len = p.size();
        for (int i = 0; i < len; i++)
        {
            for (int j = 0; j < len; j++)
            {
                g[i][j] = -dis(p[i].first, p[i].second, h[j].first, h[j].second);
                //printf("%d
", g[i][j]);
            }
        }
        nx = ny = len;
        printf("%d
", -KM());
    }
}