描述
Consider the following two-player game played with a sequence of N positive integers (2 <= N <= 100) laid onto a 1 x N game board. Player 1 starts the game. The players move alternately by selecting a number from either the left or the right end of the gameboar. That number is then deleted from the board, and its value is added to the score of the player who selected it. A player wins if his sum is greater than his opponents.
Write a program that implements the optimal strategy. The optimal strategy yields maximum points when playing against the "best possible" opponent. Your program must further implement an optimal strategy for player 2.
输入
Line 1: N, the size of the board
Line 2-etc: N integers in the range (1..200) that are the contents of the game board, from left to right
输出
Two space-separated integers on a line: the score of Player 1 followed by the score of Player 2.
样例输入
6
4 7 2 9
5 2
样例输出
18 11
题意
1*N的游戏盘，每个格子都有价值，玩家一次拿最左或最右，拿了删掉这个格子，问玩家1先拿，玩家2后拿，问俩人都最优可以拿多少分。
题解
贪心一下给否了，想了下dp
dp[i][j]=区间[i,j]先手-后手的差值
很容易推出dp[i][j]可由小区间得到
dp[i][j]=a[i]-dp[i+1][j];
dp[i][j]=a[j]-dp[i][j-1];
最终我们得到dp[1][n]为先手-后手的差值
设玩家1拿了V1,玩家2拿了V2
V1+V2=Σa
V1-V2=dp[1][n]
求得V1=(Σa+dp[1][n])/2,V2=(Σa-dp[1][n])/2
代码
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 int main() 5 { 6 int n; 7 while(scanf("%d",&n)!=EOF) 8 { 9 int a[105],dp[105][105]={0},sum=0; 10 for(int i=1;i<=n;i++) 11 scanf("%d",&a[i]),sum+=a[i]; 12 for(int len=1;len<=n;len++) 13 for(int i=1;i<=n-len+1;i++) 14 { 15 int j=i+len-1; 16 dp[i][j]=max(a[i]-dp[i+1][j],a[j]-dp[i][j-1]); 17 } 18 printf("%d %d\n",(sum+dp[1][n])/2,(sum-dp[1][n])/2); 19 } 20 return 0; 21 }