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

2015-08-18 18:32 573 查看
http://codeforces.com/gym/100650





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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