Uva 10891 sum 游戏（及其变型）；动态规划

最新推荐文章于 2020-03-10 22:59:51 发布

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最新推荐文章于 2020-03-10 22:59:51 发布

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分类专栏： ACM

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本文介绍了一种两人博弈游戏——游戏求和问题的算法实现。玩家A和B轮流从数组两端取走正整数，目标是使自己的数字总和最大化。文章提供了两种情况的代码示例：一种允许玩家连续取走多个数，另一种限制每次只能取一个数。

题目地址http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1832

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 -10 -20 7

1 2 3 4

10

代码如下

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int A[110],s[110],vis[110][110],d[110][110];

int dp(int i,int j)
{
    if(vis[i][j])
        return d[i][j];
    vis[i][j] = 1;
    int sum = s[j] - s[i-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] = sum - m;
    return d[i][j];
}

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

如果每回只能取一个数

则代码为

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int A[110],s[110],vis[110][110],d[110][110];

int dp(int i,int j)
{
    if(vis[i][j])
        return d[i][j];
    vis[i][j] = 1;
    int sum = s[j] - s[i-1];
    if(i == j)
        return A[i];
    int m;
    //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));
    m = min(dp(i+1,j),dp(i,j-1));
    d[i][j] = sum - m;
    return d[i][j];
}

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