首页 > 代码库 > POJ 2738 Two Ends(记忆化）
POJ 2738 Two Ends(记忆化）
Description In the two-player game "Two Ends", an even number of cards is laid out in a row. On each card, face up, is written a positive integer. Players take turns removing a card from either end of the row and placing the card in their pile. The player whose cards add up to the highest number wins the game. Now one strategy is to simply pick the card at the end that is the largest -- we‘ll call this the greedy strategy. However, this is not always optimal, as the following example shows: (The first player would win if she would first pick the 3 instead of the 4.) 3 2 10 4 You are to determine exactly how bad the greedy strategy is for different games when the second player uses it but the first player is free to use any strategy she wishes. Input There will be multiple test cases. Each test case will be contained on one line. Each line will start with an even integer n followed by n positive integers. A value of n = 0 indicates end of input. You may assume that n is no more than 1000. Furthermore, you may assume that the sum of the numbers in the list does not exceed 1,000,000. Output For each test case you should print one line of output of the form: In game m, the greedy strategy might lose by as many as p points. where m is the number of the game (starting at game 1) and p is the maximum possible difference between the first player‘s score and second player‘s score when the second player uses the greedy strategy. When employing the greedy strategy, always take the larger end. If there is a tie, remove the left end. Sample Input 4 3 2 10 4 8 1 2 3 4 5 6 7 8 8 2 2 1 5 3 8 7 3 0 Sample Output In game 1, the greedy strategy might lose by as many as 7 points. In game 2, the greedy strategy might lose by as many as 4 points. In game 3, the greedy strategy might lose by as many as 5 points. Source East Central North America 2005 |
上次比赛写了个记忆化，这次算是跪了。參照：点击打开链接
题意：第二个人是贪心拿法，仅仅能从两端拿，问最大的差值。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
using namespace std;
const int maxn=1100;
int dp[maxn][maxn];
int a[maxn],n;
int solve(int x,int y)
{
if(dp[x][y]!=-1)
return dp[x][y];
if(x+1==y)
return dp[x][y]=abs(a[x]-a[y]);
int sa,sb;
if(a[x+1]>=a[y])//第一个人选左端
sa=solve(x+2,y)+a[x]-a[x+1];
else
sa=solve(x+1,y-1)+a[x]-a[y];
if(a[x]<a[y-1])//第一个人选右端
sb=solve(x,y-2)+a[y]-a[y-1];
else
sb=solve(x+1,y-1)+a[y]-a[x];
return dp[x][y]=max(sa,sb);
}
int main()
{
int l=0;
while(~scanf("%d",&n)&&n)
{
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
memset(dp,-1,sizeof(dp));
printf("In game %d, the greedy strategy might lose by as many as %d points.\n",++l,solve(0,n-1));
}
return 0;
}
POJ 2738 Two Ends(记忆化）
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。