动态规划问题，每一步都分两种情况，然后选出最合适的那种

#include "iostream"
#include "memory.h"
#include "stdio.h"
using namespace std;

const int MAX = 1005;
int dp[MAX][MAX];
int largest;
int arr[MAX];

int call(int start, int end){
	if (start >= end)
		return 0;
	if (dp[start][end] != -1) //看先前有没算过这一步，如果有就直接返回
		return dp[start][end];
	else {
		int left_sum = 0, right_sum = 0;
		//player1 选择左边
		if (arr[start + 1] >= arr[end]){  //player2 选择左边
			left_sum = arr[start] - arr[start + 1] + call(start + 2, end);
		}
		else {
			left_sum = arr[start] - arr[end] + call(start + 1, end - 1);
		}
		//player1 选择右边
		if (arr[start] >= arr[end - 1]){  //player2 选择右边
			right_sum = arr[end] - arr[start] + call(start + 1, end - 1);
		}
		else {
			right_sum = arr[end] - arr[end - 1] + call(start, end - 2);
		}
		dp[start][end] = left_sum > right_sum ? left_sum : right_sum; //现在算dp
		return dp[start][end];
	}
}
int main(){
	int n;
	int gameId = 1;
	while (cin >> n && n){
		memset(dp, -1, sizeof(dp));
		memset(arr, 0, sizeof(arr));
		largest = 0;
		for (int i = 1; i <= n; i++){
			cin >> arr[i];
		}
		int sum = call(1, n);
		printf("In game %d, the greedy strategy might lose by as many as %d points.\n", gameId ++, sum);
	}
	return 0;
}

sicily 1176 two ends