ZOJ 1524 Supermarket

题意:给出一张要买的东西列表,一排货架,要求在列表上从上到下,在货架上从左到右的选购东西,如果列表上第i件物品在第j个货架上买了,那么第i+1个物品,必须在第j个货架后面买.

求买完所有东西需要的最少价钱.

设dp[i][j]为前i件物品在前j个货架上买需要的最少价钱.

那么有dp[i][j] = min(dp[i - 1][j - 1] + price[j], dp[i][j - 1]).

方程的意思是 min(前i-1件物品在前j-1个货架上买,第i件物品在第j个货架上买, i件物品都在前j-1个货架上买)　这里前提是tag[j]==list[i]　就是j货架上的是第i件物品

注意边界条件:dp[0][i] = 0

用了滚动数组优化空间.

#include <cstdio>
#include <memory.h>
#include <algorithm>
using namespace std;
const int MAX = 100010;
const int INF = 0x6FFFFFFF;

//#define min(a, b) ((a) < (b) ? (a) : (b))


int main(int argc, char const *argv[]){
	int tag[MAX];
	float price[MAX];
	float dp[MAX], tmp[MAX];
	int list[101];
	int M, N;
	while(scanf("%d%d", &M, &N) == 2){
		if(!M && !N)break;
		for(int i = 1; i <= M; ++i){
			scanf("%d", &list[i]);
		}
		for(int i = 1; i <= N; ++i){
			scanf("%d%f", &tag[i], &price[i]);
		}
		for(int i = 1; i <= N; ++i){
			dp[i] = 0;
		}
		for(int i = 1; i <= M; ++i){
			for(int j = 1; j <= N; ++j){
				tmp[j] = INF;
			}
			for(int j = i; j <= N; ++j){
				if(tag[j] == list[i] && dp[j - 1] != INF){
					if(j == 1){
						tmp[j] = dp[j - 1] + price[j];
					}else{
						tmp[j] = min(dp[j - 1] + price[j], tmp[j - 1]);
					}
				}else if(j != 1 && tmp[j - 1] != INF){
					tmp[j] = tmp[j - 1];
				}
			}
			memcpy(dp, tmp, sizeof(dp));
		}
		
		if(dp[N] != INF)
			printf("%.2lf\n", dp[N]);
		else
			printf("Impossible\n");
	}
	return 0;
}

ZOJ 1524 Supermarket,布布扣,bubuko.com

ZOJ 1524 Supermarket

原文：http://blog.csdn.net/zxjcarrot/article/details/22279879