POJ

题目：有n个数，有一个区间[a,b],第一个人先取一个数，必须在区间内，后一次取必须比第一个数大，而且差值在区间内。问最后两个人取的数的和的差值最大为多少。

思路：枚举第一次选的数，score[i]表示先手选第i个数的最大差值

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3f//0x3f3f3f3fconst int maxn=1e4+50;int num[maxn],score[maxn];int n,a,b;int dfs(int i){    if(score[i]!=-INF)        return score[i];    int mx=-INF;    for(int j=i+1;j<n;j++)        if(num[j]-num[i]>=a&&num[j]-num[i]<=b)            mx=max(mx,dfs(j));    if(mx==-INF) mx=0;    score[i]=num[i]-mx;    return score[i];}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d%d%d",&n,&a,&b);        for(int i=0;i<n;i++){            scanf("%d",&num[i]);            score[i]=-INF;        }        sort(num,num+n);        int ans=-INF;        for(int i=0;i<n;i++)            if(num[i]>=a&&num[i]<=b)                ans=max(ans,dfs(i));        if(ans==-INF) ans=0;        printf("%d\n",ans);    }    return 0;}