2005-2006 ACM-ICPC East Central North America Regional Contest (ECNA 2005) H.Two Ends

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http://codeforces.com/gym/100650

这里写图片描述

分析：
比较水的题目。
给你一个序列，有两个人分别拿，每次只能拿最左或最右，其中一个是只会贪心的拿，问贪心的人最多能输几分。
写个DP搞两边就行了，给的序列是偶数个的（题意），所以2个2个搞，只有四种情况。

#include <iostream>#include <sstream>#include <iomanip>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <bitset>#include <string>#include <numeric>#include <algorithm>#include <functional>#include <iterator>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <cctype>#include <complex>#include <ctime>typedef long long LL;const double pi = acos(-1.0);const long long mod = 1e9 + 7;using namespace std;int a[1005];int dp[1005][1005];int ans(int A,int B){    if(dp[A][B] != 0)        return dp[A][B];    if(B - A == 1)        return dp[A][B] = abs(a[A] - a[B]);    int MAX1,MAX2;    if(a[A + 1] >= a[B])        MAX1 = ans(A + 2,B) + a[A] - a[A + 1];    else        MAX1 = ans(A + 1,B - 1) + a[A] - a[B];    if(a[A] < a[B - 1])        MAX2 = ans(A,B - 2) + a[B] - a[B - 1];    else        MAX2 = ans(A + 1,B - 1) + a[B] - a[A];    return dp[A][B] = max(MAX1,MAX2);}int main(){    //freopen("int.txt","r",stdin);    //freopen("out.txt","w",stdout);    int N;    int t = 0;    while(scanf("%d",&N) && N)    {        for(int i = 1;i <= N;i++)            scanf("%d",&a[i]);        memset(dp,0,sizeof(dp));        printf("In game %d, the greedy strategy might lose by as many as %d points.\n",++t,ans(1,N));    }    return 0;}

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