POJ 2195 Going Home

时间：2019-04-08 21:10:29 阅读：157 评论：0 收藏：0 [点我收藏+]

标签：clu == while comm integer span san lin scan

Description

On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, until he enters a house. The task is complicated with the restriction that each house can accommodate only one little man.

Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a ‘.‘ means an empty space, an ‘H‘ represents a house on that point, and am ‘m‘ indicates there is a little man on that point.
技术图片

You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.

Input

There are one or more test cases in the input. Each case starts with a line giving two integers N and M, where N is the number of rows of the map, and M is the number of columns. The rest of the input will be N lines describing the map. You may assume both N and M are between 2 and 100, inclusive. There will be the same number of ‘H‘s and ‘m‘s on the map; and there will be at most 100 houses. Input will terminate with 0 0 for N and M.

Output

For each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.

Sample Input

2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0

Sample Output

2
10
28

代码：

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
using namespace std;

const int inf = 0x3f3f3f3f;
const int maxn = 1010;
int N, M;
int l1[maxn], l2[maxn], s[maxn], t[maxn];
int match1[maxn], match2[maxn];
int mp[maxn][maxn];
char maze[maxn][maxn];
int nx = 0, ny = 0;

struct Node {
    int x;
    int y;
}nodex[maxn], nodey[maxn];

int KM(int M, int N) {
    int p, q, k, i, j;
    int res = 0;
    for( i = 0; i < M; i ++) {
        l1[i] = -inf;
        for( j = 0; j < N; j ++)
            if(mp[i][j] > l1[i])
                l1[i] = mp[i][j];
        if(l1[i] == -inf) return -1;
    }
    memset(l2, 0, sizeof(l2));
    memset(match1, -1, sizeof(match1));
    memset(match2, -1, sizeof(match2));
    for( i = 0; i < M; i ++) {
        memset(t, -1, sizeof(t));
        for(s[p = q = 0] = i; p <= q && match1[i] < 0; p ++) {
            for(k = s[p], j = 0; j < N && match1[i] < 0; j ++) {
                if(l1[k] + l2[j] == mp[k][j] && t[j] < 0) {
                    s[++ q] = match2[j], t[j] = k;
                    if(s[q] < 0)
                        for(p = j; p >= 0; j = p)
                            match2[j] = k = t[j], p = match1[k], match1[k] = j;
                }
            }
        }
        if(match1[i] < 0) {
            for(i --, p = inf, k = 0; k <= q; k ++)
                for( j = 0; j < N; j ++)
                if(t[j] < 0 && l1[s[k]] + l2[j] - mp[s[k]][j] < p)
                    p = l1[s[k]] + l2[j] - mp[s[k]][j];

            for( j = 0; j < N; j ++)
            if(t[j] > 0) l2[j] += p;
            for(k = 0; k <= q; k ++)
                l1[s[k]] -= p;
        }
    }
    for( i = 0; i <M; i ++) {
        if(match1[i] < 0) return -1;
        if(mp[i][match1[i]] <= -inf) return -1;
        res += mp[i][match1[i]];
    }
    return res;
}

int main() {
    while(~scanf("%d%d", &N, &M)) {
        nx = 0, ny = 0;
        memset(mp, 0, sizeof(mp));
        if(!N && !M) break;
        for(int i = 0; i < N; i ++)
            scanf("%s", maze[i]);

        for(int i = 0; i < N; i ++) {
            for(int j = 0; j < M; j ++) {
                if(maze[i][j] == ‘m‘) {
                    nodex[nx].x = i;
                    nodex[nx].y = j;
                    nx ++;
                } else if(maze[i][j] == ‘H‘) {
                    nodey[ny].x = i;
                    nodey[ny].y = j;
                    ny ++;
                }
            }
        }

        for(int i = 0; i < nx; i ++) {
            for(int j = 0; j < ny; j ++) {
                mp[i][j] = -abs(nodex[i].x - nodey[j].x) - abs(nodex[i].y - nodey[j].y);
            }
        }
        printf("%d\n", -KM(nx, ny));
    }
    return 0;
}

　　KM 求最小权匹配

刚刚结束携程的笔试最后一道编程题看着乱糟糟样例打表居然能过 85.7% 数据可以说是很水了

POJ 2195 Going Home

标签：clu == while comm integer span san lin scan

原文地址：https://www.cnblogs.com/zlrrrr/p/10673255.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行