poj1678 I Love this Game!

I Love this Game!

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 1560 Accepted: 602

Description

A traditional game is played between two players on a pool of n numbers (not necessarily distinguishing ones).

The first player will choose from the pool a number x1 lying in [a, b] (0 < a < b), which means a <= x1 <= b. Next the second player should choose a number y1 such that y1 - x1 lies in [a, b] (Attention! This implies y1 > x1 since a > 0). Then the first player should choose a number x2 such that x2 - y1 lies in [a, b]... The game ends when one of them cannot make a choice. Note that a player MUST NOT skip his turn.

A player's score is determined by the numbers he has chose, by the way:

player1score = x1 + x2 + ...
player2score = y1 + y2 + ...

If you are player1, what is the maximum score difference (player1score - player2score) you can get? It is assumed that player2 plays perfectly.

Input

The first line contains a single integer t (1 <= t <= 20) indicating the number of test cases. Then follow the t cases. Each case contains exactly two lines. The first line contains three integers, n, a, b (2 <= n <= 10000, 0 < a < b <= 100); the second line contains n integers, the numbers in the pool, any of which lies in [-9999, 9999].

Output

For each case, print the maximum score difference player1 can get. Note that it can be a negative, which means player1 cannot win if player2 plays perfectly.

Sample Input

36 1 21 3 -2 5 -3 62 1 2-2 -12 1 21 0

Sample Output

-301

记义化搜索，因为，我们用dp[i]表示先手搜到第i个是，能得到的最大值，那么对于任意状态都用dp[i]=pri[i]-max{dp[j]};相减一下，刚好就是，把先手后手交换了，可以用一个式子，统一表示，j是i的后继结点，这样，就可以得出结果了！

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define M 10050#define inf 0x4f4f4f4fint pri[M],a,b,n,dp[M];int dfs(int now){    if(dp[now]!=-inf)return dp[now];    int ans=-inf,k;    for(int i=now+1;i<n;i++){        int temp=pri[i]-pri[now];        if(temp>b)break;        if(temp>=a&&(k=dfs(i))>ans){            ans=k;        }    }    if(ans==-inf)ans=0;    return dp[now]=pri[now]-ans;}int main(){    int tcase,i,j,k;    scanf("%d",&tcase);    while(tcase--){        scanf("%d%d%d",&n,&a,&b);        for(i=0;i<n;i++)        dp[i]=-inf;        for(i=0;i<n;i++)        scanf("%d",&pri[i]);        sort(pri,pri+n);        for(i=1,j=0;i<n;i++){            if(pri[i]!=pri[j])pri[++j]=pri[i];        }        int ans=-inf;        for(i=0;i<n;i++){            if(pri[i]<=b&&pri[i]>=a&&(k=dfs(i))>=ans){                ans=k;            }        }        if(ans==-inf)printf("0\n");        else printf("%d\n",ans);    }    return 0;}