TYVJ P1047 - 乘积最大(DP)

题意:给出一个数字串,把它划分成K + 1份,使其乘积最大.

类似矩阵连乘的DP 设dp[i][j][p]为区间[i, j]划分成p份的最大乘积.

那么dp[i][j][p] = max(dp[i][k][p - 1] * val[k + 1][j]) val[i][j]为数字串i到j的值.

base case: dp[i][j][1] = val[i][j]

#include <cstdio>
#include <memory.h>
#include <algorithm>
using namespace std;
const int MAX = 41;

double dp[MAX][MAX][7];
char num[MAX];
double val[MAX][MAX];
int main(int argc, char const *argv[]){
	int N, K;
	scanf("%d%d", &N, &K);
	scanf("%s", num + 1);

	for(int i = 1; i <= N; ++i){
		val[i][i] = num[i] - ‘0‘;
		for(int j = i + 1; j <= N; ++j){
			val[i][j] = val[i][j - 1] * 10 + num[j] - ‘0‘;
			dp[i][j][1] = val[i][j];
		}
		dp[i][i][1] = val[i][i];
	}
	
	++K;
	for(int m = 1; m < N; ++m){
		for(int i = 1; i < N && i + m <= N; ++i){
			int j = i + m;
			for(int p = 2; p <= K && p <= i + m; ++p){
				for(int k = i; k < j; ++k){
					dp[i][j][p] = max(dp[i][j][p], dp[i][k][p - 1] * val[k + 1][j]);
				}
			}
		}
	}
	printf("%.0lf\n", dp[1][N][K]);
	return 0;
}

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TYVJ P1047 - 乘积最大(DP)

原文：http://blog.csdn.net/zxjcarrot/article/details/22508591