HDU 6199 gems gems gems (2017沈阳网赛

题意：

有一堆数， 两个人轮流取， 只能从最左边开始选择， 假设上一个人选了k 个牌， 那么下一个人只能选择k 或 k + 1 张牌。 第一个人得分为A， 第二个人得分为B， 求A- B， 每个人的策略都想使自己得分尽可能的高。

思路：

是UVA 10891 的变形把。

我们令dp[i][j] 表示从i 位置开始选择， 能选j 个牌的最大分数差值。

那么直接根据j 来转移即可。

总共两种选择， 要么选j 个， 要么选j + 1个， 处理个前缀和即可。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 20000 + 1;const int inf = 0x3f3f3f3f;int dp[maxn][200];bool vis[maxn][200];int a[maxn], n;int sum[maxn], ks;int dfs(int pos, int k){    int& ans = dp[pos][k];    if (vis[pos][k] == 1) return ans;    vis[pos][k] = 1;    if (pos > n){        return ans = 0;    }    if (pos + k - 1 > n){        return ans = 0;    }    ans = -inf;    int lak = pos + k - 1;    int lak1 = pos + k;    if (lak <= n){        ans = max(ans, sum[lak] - sum[pos - 1] - dfs(lak + 1, k));    }    if (lak1 <= n){        ans = max(ans, sum[lak+1] - sum[pos - 1] - dfs(lak1 + 1, k + 1));    }    return ans;}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d", &n);        memset(vis,0,sizeof vis);        for (int i = 1; i <= n; ++i){            scanf("%d", &a[i]);            sum[i] = sum[i-1] + a[i];        }        printf("%d\n", dfs(1, 1));    }    return 0;}

gems gems gems

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1296    Accepted Submission(s): 276

Problem Description

Now there are n gems, each of which has its own value. Alice and Bob play a game with these n gems.
They place the gems in a row and decide to take turns to take gems from left to right. 
Alice goes first and takes 1 or 2 gems from the left. After that, on each turn a player can take k or k+1 gems if the other player takes k gems in the previous turn. The game ends when there are no gems left or the current player can't take k or k+1 gems.
Your task is to determine the difference between the total value of gems Alice took and Bob took. Assume both players play optimally. Alice wants to maximize the difference while Bob wants to minimize it.

 

Input

The first line contains an integer T (1≤T≤10), the number of the test cases. 
For each test case:
the first line contains a numbers n (1≤n≤20000);
the second line contains n numbers: V1,V2…Vn. (−100000≤Vi≤100000)

 

Output

For each test case, print a single number in a line: the difference between the total value of gems Alice took and the total value of gems Bob took.

 

Sample Input

131 3 2

 

Sample Output

4

 

Source

2017 ACM/ICPC Asia Regional Shenyang Online

 

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