hihocoder-Week173--A Game?

A Game

時(shí)間限制:10000ms 單點(diǎn)時(shí)限:1000ms 內(nèi)存限制:256MB

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(qū)間dp，

?

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

應(yīng)該對(duì)方的策略不是讓我方最少，而是對(duì)方也取得最優(yōu)。

?

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(yōu)，所以 dp[i][j] 指的是當(dāng)前下手的選手，可以取得的最優(yōu)成果。 所以當(dāng)前狀態(tài)是依賴于前面的 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; }

　　

?

轉(zhuǎn)載于:https://www.cnblogs.com/zhang-yd/p/7718448.html

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