Martian Mining poj2948

这题要注意决策的转移,对每一行来说,选一个点作为基准点k,基准点左边(j < k)的点全部向左运输,基准点向右(j >= k)全部向上运输,不难想到其他的决策都不可能得到更优解,

而够转移到该状态的上一行状态的基准点一定是(j >= k),如果上一行基准点j<k的话则 j<= i < k的矿是无法向上运输的


#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstring>
#include <stack>
#include <cctype>
#include <utility>   
#include <map>
#include <string>  
#include <climits> 
#include <set>
#include <string> 
#include <sstream>
#include <utility>
#include <ctime>
 
using std::priority_queue;
using std::vector;
using std::swap;
using std::stack;
using std::sort;
using std::max;
using std::min;
using std::pair;
using std::map;
using std::string;
using std::cin;
using std::cout;
using std::set;
using std::queue;
using std::string;
using std::istringstream;
using std::make_pair;
using std::greater;
using std::endl;

const int MAXN(510);

int table[MAXN], mx[MAXN];
int A[MAXN][MAXN], B[MAXN][MAXN];

int main()
{
	int n, m;
	while(scanf("%d%d", &n, &m), n || m)
	{
		for(int i = 1; i <= n; ++i)
			for(int j = 1; j <= m; ++j)
				scanf("%d", A[i]+j);
		for(int i = 1; i <= n; ++i)
			for(int j = 1; j <= m; ++j)
				scanf("%d", B[i]+j);
		memset(mx, 0, sizeof(mx));
		for(int i = n; i >= 1; --i)
		{
			int suma = 0, sumb = 0;
			for(int j = 1; j <= m; ++j)
				sumb += B[i][j];
			for(int j = 1; j <= m+1; ++j)
			{
				table[j] = mx[j]+suma+sumb;
				suma += A[i][j];
				sumb -= B[i][j];
			}
			mx[m+1] = table[m+1];
			for(int j = m; j >= 1; --j)
				mx[j] = max(table[j], mx[j+1]);
		}
		int ans = 0;
		for(int i = 1; i <= m+1; ++i)
			ans = max(ans, table[i]);
		printf("%d\n", ans);
	}
	return 0;
}


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