uva 10891 Game of Sum (DP)

uva 10891 Game of Sum (DP)

This is a two player game. Initially there are n integer numbers in an array and players
A and B get chance to take them alternatively. Each player can take one or more numbers from the left or right end of the array but cannot take from both ends at a time. He can take as many consecutive numbers as he wants during
his time. The game ends when all numbers are taken from the array by the players. The point of each player is calculated by the summation of the numbers, which he has taken. Each player tries to achieve more points from other. If both players play optimally
and player A starts the game then how much more point can player
A get than player B?

Input

The input consists of a number of cases. Each case starts with a line specifying the integer
n (0 < n ≤100), the number of elements in the array. After that,
n numbers are given for the game. Input is terminated by a line where
n=0.



Output

For each test case, print a number, which represents the maximum difference that the first player obtained after playing this game optimally.

Sample Input Output for Sample Input

4 4 -10 -20 7 4 1 2 3 4 0

7 10

题目大意：有n个石头排成一条排，然后有两个人来玩游戏， 每个人每次可以从两端（左或右）中的任意一端取走若干个石头（获得价值为取走石头之和）， 但是他取走的方式一定要让他在游戏结束时价值尽量的高，两个人都很聪明，所以每一轮两人都将按照对自己最有利的方法去取数字，请你算一下在游戏结束时，先取数的人价值与后取数人价值之差。

解题思路：dp函数返回的是i到j玩家1可以取得最大值。

#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
using namespace std;
int S[105], A[105], d[105][105], vis[105][105], n;
int dp(int i, int j) { //dp函数返回的是i到j玩家1可以取得最大值
	if (vis[i][j]) return d[i][j];
	vis[i][j] = 1;
	int m = 0;
	for (int k = i + 1; k <= j; k++) {
		m = min(m, dp(k, j));
	}
	for (int k = i; k < j; k++) {
		m = min(m, dp(i, k));
	}
	d[i][j] = S[j] - S[i - 1] - m;
	return d[i][j];
}
int main() {
	while (scanf("%d", &n), n) {
		memset(vis, 0, sizeof(vis));
		S[0] = 0;
		for (int i = 1; i <= n; i++) {
			scanf("%d", &A[i]);
			S[i] = S[i - 1] + A[i];
		}
		printf("%d\n", 2 * dp(1, n) - S
); //dp(1, n) - (S
 - dp(1, n))
	}	
	return 0;
}