解題區/解題區 - UVa

UVa 12637 - 30 Minutes or Less

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

現在有 R 個 Pizza 店，同一個時間接收到 N 筆訂單，每一個店負責處理一筆訂單，發送的成本為店家到訂單的曼哈頓距離。

求所有訂單送達的總時間最小值為何？

Sample Input

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7
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11
2 2
1 5
2 1
4 2
4 3
3 2
2 1
4 3
7 4
4 5
5 -1

Sample Output

1
2
8
7

Solution

一開始以為是要求最後一個送達的時間最小化，二分下去才發現連範測都過不了。

建圖之後，用 KM 算法找最大帶權匹配 (費用最小就取負號)，用最少費用流解也是可以。

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#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <queue>
#include <stack>
#include <assert.h>
#include <set>
#include <map>
#include <vector>
using namespace std;
struct KM {
    int W[105][105], N;
    int mx[105], my[105]; // match arr
    int lx[105], ly[105]; // label arr
    int x[105], y[105]; // used arr
    int hungary(int nd) {
        int i;
        x[nd] = 1;
        for(i = 0; i < N; i++) {
            if(y[i] == 0 && W[nd][i] == lx[nd]+ly[i]) {
                y[i] = 1;
                if(my[i] == -1 || hungary(my[i])) {
                    my[i] = nd;
                    return 1;
                }
            }
        }
        return 0;
    }
    int run() {
        int i, j, k, d;
        memset(mx, -1, sizeof(mx));
        memset(my, -1, sizeof(my));
        for (int i = 0; i < N; i++)
        	lx[i] = -0x3f3f3f3f, ly[i] = 0;
        for(i = 0; i < N; i++)
            for(j = 0; j < N; j++)
                lx[i] = lx[i] > W[i][j] ? lx[i] : W[i][j];
        for(i = 0; i < N; i++) {
            while(1) {
                memset(x, 0, sizeof(x));
                memset(y, 0, sizeof(y));
                if(hungary(i))  break;
                d = 0x3f3f3f3f;
                for(j = 0; j < N; j++) {
                    if(x[j]) {
                        for(k = 0; k < N; k++)
                            if(!y[k])
                            d = d < lx[j]+ly[k]-W[j][k] ?
                                d : lx[j]+ly[k]-W[j][k];
                    }
                }
                if(d == 0x3f3f3f3f)  break;
                for(j = 0; j < N; j++) {
                    if(x[j])    lx[j] -= d;
                    if(y[j])    ly[j] += d;
                }
            }
        }
        int res = 0;
        for(i = 0; i < N; i++) {
            if(my[i] != -1)
                res += W[my[i]][i];
        }
        return res;
    }
} km;
int dist(int x1, int y1, int x2, int y2) {
    return abs(x1 - x2) + abs(y1 - y2);
}
int main() {
    int N, R;
    int x[256], y[256], cost[128][128];
    while (scanf("%d %d", &R, &N) == 2) {
        for (int i = 0; i < R; i++)
            scanf("%d %d", &x[i], &y[i]);
        for (int i = 0; i < N; i++)
            scanf("%d %d", &x[i+R], &y[i+R]);
        
        km.N = R;
        for (int i = 0; i < R; i++) {
            for (int j = 0; j < N; j++) {
                cost[i][j] = dist(x[i], y[i], x[j+R], y[j+R]);
                km.W[i][j] = -cost[i][j];
            }
            for (int j = N; j < R; j++)
                km.W[i][j] = 0;
        }
        int ret = km.run();
        printf("%d\n", -ret);
    }
    return 0;
} 
/*
2 2
1 5
2 1
4 2
4 3
3 2
2 1
4 3
7 4
4 5
5 -1
*/