ZOJ 1428 Magazine Delivery(DP)

题意:有一堆杂志要送到N个地点,有３两出租车可以用,每辆杂志可以载无限多的杂志,初始３两车都在１地点,两地点距离由D[i][j]给出,第i个地点只有第i- 1个地点送了杂志之后才能送,

求使用３两车送完杂志的最少距离.

思路:题目中的条件:第i个地点只有第i- 1个地点送了杂志之后才能送　给出了很好的一个阶段划分.

设dp[i][j][k][m]为送第i个地点的时候３两车分别在j,k,m位置

那么dp[i][j][k][m] = min(dp[i - 1][j][m][k] + D[j][i], dp[i - 1][j][m][k] + D[k][i], dp[i - 1][j][m][k] + D[m][i])

base case: dp[1][1][1][1] = 0

#include <cstdio>
#include <memory.h>
#include <algorithm>
using namespace std;
const int MAX = 31;


int main(int argc, char const *argv[]){
	int dp[MAX][MAX][MAX][MAX];
	int D[MAX][MAX];
	int M, N;
	scanf("%d", &M);
	while(M--){
		scanf("%d", &N);
		for(int i = 1; i < N; ++i){
			for(int j = i + 1; j <= N; ++j){
				int v;
				scanf("%d", &v);
				D[i][j] = D[j][i] = v;
			}
		}
		memset(dp, 0x6f, sizeof(dp));
		dp[1][1][1][1] = 0;
		for(int i = 2; i <= N; ++i){
			for(int j = 1; j <= N; ++j){
				for(int k = 1; k <= N; ++k){
					for(int m = 1; m <= N; ++m){
						if(dp[i - 1][j][k][m] == 0x6f6f6f6f)continue;
						dp[i][i][k][m] = min(dp[i][i][k][m], dp[i - 1][j][k][m] + D[j][i]);
						dp[i][j][i][m] = min(dp[i][j][i][m], dp[i - 1][j][k][m] + D[k][i]);
						dp[i][j][k][i] = min(dp[i][j][k][i], dp[i - 1][j][k][m] + D[m][i]);
					}
				}
			}
		}
		printf("%d\n", *min_element(&dp[N][0][0][0], &dp[N][N][N][N]));
	}
	return 0;
}

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ZOJ 1428 Magazine Delivery(DP)

原文：http://blog.csdn.net/zxjcarrot/article/details/22895007