UVA - 10891 Game of Sum

Game of Sum

Time Limit: 3000MSMemory Limit: Unknown64bit IO Format: %lld & %llu

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Description

Problem E
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, n numbers 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

7

10

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Problem setter: Syed Monowar Hossain

Special Thanks: Derek Kisman, Mohammad Sajjad Hossain

分析：

解法一：

刘汝佳大白书的dp题，同样不好想。采用了记忆化搜索。时间复杂度O（n^3）,空间复杂度O(n^2)

ac代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;


const int maxn=100+10;
int s[maxn],a[maxn],d[maxn][maxn],vis[maxn][maxn],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));
    for(int k=i;k<j;k++)
    m=min(m,dp(i,k));
    d[i][j]=s[j]-s[i-1]-m;
    return d[i][j];
}


int main()
{
    while(scanf("%d",&n)&&n)
    {
        s[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            s[i]=s[i-1]+a[i];
        }
        memset(vis,0,sizeof(vis));
        printf("%d\n",2*dp(1,n)-s[n]);
    }
    return 0;
}

 

解法二：

进一步优化，时间复杂度O（n^2）

ac代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;


const int maxn=100+10;
int s[maxn],a[maxn],d[maxn][maxn],n,f[maxn][maxn],g[maxn][maxn];


int main()
{
    while(scanf("%d",&n)&&n)
    {
        s[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            s[i]=s[i-1]+a[i];
        }
        for(int i=1;i<=n;i++)
        f[i][i]=g[i][i]=d[i][i]=a[i];
        for(int L=1;L<n;L++)
        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]=s[j]-s[i-1]-m;
            f[i][j]=min(d[i][j],f[i+1][j]);
            g[i][j]=min(d[i][j],g[i][j-1]);
        }
        printf("%d\n",2*d[1][n]-s[n]);
    }
    return 0;
}