UVA 10891 Game of Sum dp（记忆化搜索）

博弈论游戏DP算法解析

最新推荐文章于 2020-10-12 21:14:59 发布

原创最新推荐文章于 2020-10-12 21:14:59 发布 · 698 阅读

0 ·

CC 4.0 BY-SA版权

ACM 同时被 2 个专栏收录

20 篇文章

订阅专栏

4 篇文章

订阅专栏

本文介绍了一款两玩家博弈游戏的问题，通过动态规划解决如何让先手玩家获得尽可能多的分数。文章提供了完整的AC代码，并详细解释了状态转移方程。

Problem E

Game of Sum
Input File: e.in

Output: Standard Output

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 -10 -20 7

1 2 3 4

10

题意：A和B在博弈，A先开始。每次选手可以取左边连续n个或者右边连续n个，取到没有数字了，计算每个人所有取的数字之和为每个人的分数。A和B都按最优的选择来玩，问最后A比B多多少分。

思路：dp。dp[l][r]表示区间[l,r]之间A能得到的最大分数，状态转移方程为：dp[l][r]=sum[l][r]-min(dp[l+k][r],dp[l][r-k])

本题因为数据范围比较小，我于是用记忆化搜索来做的。写搜索总比dp写的顺……

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#define INF 10000006
using namespace std;
int f[105][105],a[105],n;
int dfs(int l,int r){
	if(l>r)return 0;
	if(f[l][r]<INF)return f[l][r];
	for(int i=1;i<=r-l+1;i++){
		int a=dfs(l,r-i);
		int b=dfs(l+i,r);
		f[l][r]=min(f[l][r],a<b?a:b);
	}
	f[l][r]=a[r]-a[l-1]-f[l][r];
	return f[l][r];
}
int main(){
	int n;
	while(scanf("%d",&n),n){
		for(int i=0;i<=n;i++)
		for(int j=0;j<=n;j++)
		f[i][j]=INF;
		f[0][0]=a[0]=0;
		for(int i=1;i<=n;i++){
			scanf("%d",a+i);
			f[i][i]=a[i];
			a[i]+=a[i-1];
		}
		printf("%d\n",2*dfs(1,n)-a[n]);
		//for(int i=1;i<=n;i++){
		//	for(int j=i;j<=n;j++){
		//		cout<<i<<' '<<j<<' '<<f[i][j]<<endl;
		//	}
		//}
	}

}