Game of Sum 记忆化搜索/递推

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, nnumbers 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

4 -10 -20 7

4

1 2 3 4

0

解决方案：可得递推公式：d[i][j]=sum[j]-sum[i-1]-min(d[i][j],d[i+1][j],.....,d[j][j],d[i][j-1],d[i][j-2],.....d[i][i],0);d[i][j]表示i--j的范围，先手能拿到最大的和。递推时，可从左往右，按第一个，拿前两个，前三，。。。，从右往左，拿后1个，后2个，。。。。，还有全拿的。0就表示全拿，这样有了隐式的边界条件，就不用显式边界条件了。代码一是用记忆化搜索的方式。代码二：把方程变形了。f[i][j]=min(d[i][j],d[i+1][j],....d[j][j]),g[i][j]=min(d[i][j],d[i][j-1],....,d[i][i])；

状态可写成d[i][j]=sum[j]-sum[i-1]-min(f[i+1][j],d[i][j-1],0).f[i][j]=min{f[i+1][j],d[i][j]},g[i][j]=min{f[i][j-1],d[i][j]}；这样时间复杂度降了很多。

7

10

code1：

#include<iostream>#include<cstdio>#include<cstdio>#include<cstring>using namespace std;int d[103][103];int sum[103];bool vis[103][103];int n;int dp(int i,int j){    if(vis[i][j]) return d[i][j];    vis[i][j]=1;    int m=0;    for(int k=i+1;k<=j;k++) m=min(m,dp(k,j));///从左边取    for(int k=i;k<j;k++) m=min(m,dp(i,k));///从右边取    d[i][j]=sum[j]-sum[i-1]-m;    return d[i][j];}///记忆化搜索int main(){    while(~scanf("%d",&n)&&n){        memset(sum,0,sizeof(sum));        for(int i=1;i<=n;i++){            int a;            scanf("%d",&a);            sum[i]=sum[i-1]+a;        }        memset(vis,false,sizeof(vis));        printf("%d\n",2*dp(1,n)-sum[n]);    }    return 0;}

code2：<pre name="code" class="cpp">#include<iostream>#include<cstdio>#include<cstdio>#include<cstring>using namespace std;int d[103][103];int sum[103];int f[103][103],g[103][103];int n,a[103];int main(){    while(~scanf("%d",&n)&&n){        memset(sum,0,sizeof(sum));        for(int i=1;i<=n;i++){            scanf("%d",&a[i]);            sum[i]=sum[i-1]+a[i];        }        for(int i=1;i<=n;i++) g[i][i]=f[i][i]=d[i][i]=a[i];///边界条件        for(int L=1;L<n;L++){///按L=j-i递增计算            for(int i=1;i+L<=n;i++){                int j=i+L;                int m=0;                m=min(m,f[i+1][j]);                m=min(m,g[i][j-1]);                d[i][j]=sum[j]-sum[i-1]-m;                f[i][j]=min(d[i][j],f[i+1][j]);///为下一次结果更新g                g[i][j]=min(d[i][j],g[i][j-1]);///为下一次结果更新f            }        }        printf("%d\n",2*d[1][n]-sum[n]);    }    return 0;}