10891 – Game of Sum
Time limit: 3.000 seconds
This is a two player game. Initially there areninteger numbers in an array and playersAandBget 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 playerAstarts the game then how much more point can playerAget than playerB?
Input
The input consists of a number of cases. Each case starts with a line specifying the integern(0 < n ≤100), the number of elements in the array. After that,nnumbers are given for the game. Input is terminated by a line wheren=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 InputOutput for Sample Input
4
4 -10 -20 7
4
1 2 3 4
0
7
10
题意：给定n个数字，A和B可以从这串数字的两端任意选数字，，一次只能从一端选取。
思路：dp[ i ][ j ] 表示先手玩家在区间 [ i , j ] 所能取得的石子数量的最大值，那么
dp[ i ][ j ] = sum[ j ] – sum[ i – 1 ] – min(dp[ i ][ k ] , dp[ k + 1 ][ j ]) ,其中sum为前缀和。
#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int inf=1<<30;const int maxn=110;int dp[maxn][maxn],a[maxn],sum[maxn],n;void initial(){for(int i=0;i<=n;i++)for(int j=0;j<=n;j++)dp[i][j]=-inf;sum[0]=0;}void input(){for(int i=1;i<=n;i++){scanf("%d",&a[i]);sum[i]=sum[i-1]+a[i];}}int DP(int l,int r){if(dp[l][r]!=-inf) return dp[l][r];if(l==r) return dp[l][r]=a[l];int ans=sum[r]-sum[l-1];for(int i=l;i<r;i++)ans=max(ans,sum[r]-sum[l-1]-min(DP(l,i),DP(i+1,r)));return dp[l][r]=ans;}int main(){while(scanf("%d",&n)!=EOF){if(n==0) break;initial();input();DP(1,n);printf("%d\n",dp[1][n]-(sum[n]-dp[1][n]));}return 0;}
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