POJ 1160 Post Office (二维DP)

题目描述

 

总结

1. [i~j] 之间建立一个邮局, 位置是 (i+j+1)/2

2. 初始化, 尽可能少的初始化

 

代码

/*
 * source.cpp
 *
 *  Created on: 2014-4-5
 *      Author: vincent
 */

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <memory.h>
using namespace std;

const int NUM = 310;
const int INF = 0X3F3F3F3F;
int arr[10010];
int dist[NUM][NUM];
int dp[NUM][31];
void init(int n) {
	for(int i = 1; i <= n; i ++) {
		for(int j = i; j <= n; j ++) {
			int mid = (i+j+1) >> 1;
			int sum = 0;
			for(int k = i; k <= j; k ++)
				sum += abs(arr[k] - arr[mid]);
			dist[i][j] = sum;
		}
	}
}

int cal(int n, int m) {
	memset(dp, 0X3F, sizeof(dp));

	for(int i = 1; i <= n; i ++) {
		for(int j = 1; j <= m; j ++) {
			if(j >= i) {
				dp[i][j] = 0;
				continue;
			}
			dp[i][j] = dist[1][i];
			//cout << dp[i][j] << endl;
			for(int k = 1; k < i; k ++) {
				dp[i][j] = min(dp[i][j], dp[k][j-1] + dist[k+1][i]);
			}
		}
	}
	return dp[n][m];
}

int main() {
	freopen("C:\\Users\\vincent\\Dropbox\\workplacce\\joj\\test.txt", "r", stdin);
	int n, m;
	while(scanf("%d%d", &n, &m) != EOF) {
		for(int i = 1; i <= n; i ++)
			scanf("%d", arr+i);
		init(n);
		int res = cal(n, m);
		printf("%d\n", res);
	}
	return 0;
}