POJ 1157

简单的dp。状态定义好就迎刃而解，令(dp(i, j))为第(i)种花必须在第j个花瓶，且满足条件的最大aesthetic value。

状态转移方程也很好想，估计了下时间复杂度发现没问题，就着手实现了

#include <iostream>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <stack>
#include <map>
#include <set>
using namespace std;

const int maxf= 105;
const int maxv= 105;
const int INF= 0x3f3f3f3f;

int fv[maxf][maxv], dv[maxf][maxv];
int f, v;

int main(int argc, char const *argv[])
{
	scanf("%d %d", &f, &v);

	for (int i= 1; i<= f; ++i){
		for (int j= 1; j<= v; ++j){
			scanf("%d", fv[i]+j);
		}
	}
	memset(dv, 0xc0, sizeof(dv));

	for (int i= 0; i<= v; ++i){
		dv[0][i]= 0;
	}
	for (int i= 1; i<= f; ++i){
		dv[i][i]= dv[i-1][i-1]+fv[i][i];
	}

	int ud= v-f;
	for (int d= 1; d<= ud; ++d){
		for (int i= 1, j= i+d; j<= v; ++i, ++j){
			int tv= dv[i][j];
			for (int k= i-1; k< j; ++k){
				tv= max(tv, dv[i-1][k]);
			}
			dv[i][j]= tv+fv[i][j];
		}
	}

	int ans= -INF;
	for (int i= f; i<= v; ++i){
		ans= max(ans, dv[f][i]);
	}

	printf("%d
", ans);
	return 0;
}