TYVJ P1203 - 机器分配

题目地址:TYVJ P1203 - 机器分配

简单DP 设dp[i][j]为前i个公司分配j个机器能得到的最大利益

dp[i][j] = max{prof[i][k] + dp[i - 1][j - k] | 0<=k<= j}

#include <cstdio>
#include <memory.h>
#include <algorithm>
using namespace std;
const int MAX = 101;

int dp[MAX][MAX];
int prof[MAX][MAX];
int main(int argc, char const *argv[]){
	int N, M;
	scanf("%d%d", &N, &M);
	for(int i = 1; i <= N; ++i){
		for(int j = 1; j <= M; ++j){
			scanf("%d", &prof[i][j]);
		}
	}
	memset(dp, 0, sizeof(dp));
	for(int i = 1; i <= N; ++i){
		for(int j = 1; j <= M; ++j){
			for(int k = 0; k <= j; ++k){
				dp[i][j] = max(dp[i][j], prof[i][k] + dp[i - 1][j - k]);
			}
		}
	}
	printf("%d\n", dp[N][M]);
	return 0;
}

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TYVJ P1203 - 机器分配

原文：http://blog.csdn.net/zxjcarrot/article/details/22485123