hihocoder-Week173--A Game

A Game

描述

Little Hi and Little Ho are playing a game. There is an integer array in front of them. They take turns (Little Ho goes first) to select a number from either the beginning or the end of the array. The number will be added to the selecter‘s score and then be removed from the array.

Given the array what is the maximum score Little Ho can get? Note that Little Hi is smart and he always uses the optimal strategy.

輸入

The first line contains an integer N denoting the length of the array. (1 ≤ N ≤ 1000)

The second line contains N integers A1, A2, ... AN, denoting the array. (-1000 ≤ Ai ≤ 1000)

輸出

Output the maximum score Little Ho can get.

樣例輸入

4
-1 0 100 2

樣例輸出

99

使用區間dp，

但是我的這種方法只ac了50%， 應該是dp[i][j][1] = max( dp[i][j][1] , min( dp[i+1][j][0] , dp[i][j-1][0])

應該對方的策略不是讓我方最少，而是對方也取得最優。

AC 50% Code:

#include <cstdio> 
#include <cstring> 

#include <iostream> 
using namespace std; 

const int MAXN = 1000 + 10; 

int n, num[MAXN], dp[MAXN][MAXN][2]; 


int main(){
	freopen("in.txt", "r", stdin); 

	int n; 
	scanf("%d", &n); 

	for(int i=1; i<=n; ++i){
		scanf("%d", &num[i]); 
	} 

	memset(dp, 0, sizeof(dp)); 

	for(int i=1; i<=n; ++i){
		dp[i][i][0] = num[i]; 
	}

	for(int i=n; i>=1; --i){
		for(int j=i; j<=n; ++j){
			dp[i][j][0] = max( dp[i+1][j][1] + num[i],   dp[i][j][0] ); 

			dp[i][j][0] = max( dp[i][j-1][1] + num[j],   dp[i][j][0] ); 
			
			dp[i][j][1] = max( dp[i][j][1], min( dp[i+1][j][0] ,  dp[i][j-1][0] ) ); 
		}
	} 
	int ans = dp[1][n][0]; 

	printf("%d\n", ans);

	return 0; 
}

所以，

dp[i][j] = max( sum(i,j) - dp[i+1][j], sum(i,j) - dp[i][j-1])

雙方都在求最優，所以 dp[i][j] 指的是當前下手的選手，可以取得的最優成果。 所以當前狀態是依賴於前面的 dp[i][j-1] 和 dp[i+1][j] ，

AC Code

#include <cstdio> 
#include <cstring> 

#include <iostream> 
using namespace std; 

const int MAXN = 1000 + 10; 

int n, num[MAXN], sum[MAXN], dp[MAXN][MAXN]; 

int main(){

	int n; 
	scanf("%d", &n); 

    sum[0] = 0; 
	for(int i=1; i<=n; ++i){
		scanf("%d", &num[i]); 
		sum[i] = sum[i-1] + num[i];  
	} 

	memset(dp, 0, sizeof(dp)); 
	
	for(int i=1; i<=n; ++i){
		dp[i][i] = num[i]; 
	}

	for(int i=n; i>=1; --i){
		for(int j=i+1; j<=n; ++j){
			dp[i][j] = (sum[j] - sum[i-1]) - min( dp[i+1][j], dp[i][j-1] ); 
		}
	} 

	int ans = dp[1][n]; 

	printf("%d\n", ans);

	return 0; 
}

hihocoder-Week173--A Game