POJ2738 Two Ends (DP)

                                                                                                                                                     Two Ends

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1811 Accepted: 855

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 48 1 2 3 4 5 6 7 88 2 2 1 5 3 8 7 30

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.

2738Accepted4080K32MSC++
1537B

题意是两个人在博弈，一个人用贪心的策略，一个人是理性的，问理性的能够拿到的数最大能比贪心的人拿到的数大多少。初看上去不像是DP，其实仔细想想就可以知道这题是一道DP。隐藏的比较深，找出它的动态方程比较难，分四种情况讨论。开始的时候我只分两种情况讨论了，郁闷了很久。开始的动态方程：

dp[i][j]=max(max(dp[i+2][j]+a[i]-a[i+1],dp[i][j-2]+a[j]-a[j-1]), max(dp[i+1][j-1]+a[i]-a[j],dp[i+1][j-1]+a[j]-a[i]));

之后想明白了  原来是四种情况，这是之后写的动态方程：

if(a[j]<=a[i+1]) dp[i][j]=max(dp[i][j],dp[i+2][j]+a[i]-a[i+1]);if(a[i]<a[j-1]) dp[i][j]=max(dp[i][j],dp[i][j-2]+a[j]-a[j-1]);if(a[i+1]<a[j]) dp[i][j]=max(dp[i][j],dp[i+1][j-1]+a[i]-a[j]);if(a[i]>=a[j-1]) dp[i][j]=max(dp[i][j],dp[i+1][j-1]+a[j]-a[i]);

还是贴上代码，大家有什么好的意见，可以和我讨论讨论：

#include <iostream>#include <cstring>#include <cmath>using namespace std;int n,a[1001],dp[1001][1001];int main(){ int t=0; while(scanf("%d",&n)&&n) { int i,j,k; for(i=1;i<=n;i++) scanf("%d",&a[i]); memset(dp,0,sizeof(dp)); for(i=1;i<n;i++) dp[i][i+1]=abs(a[i]-a[i+1]); for(k=4;k<=n;k+=2) for(i=1;i<=n-k+1;i++) { j=i+k-1; if(a[j]<=a[i+1]) dp[i][j]=max(dp[i][j],dp[i+2][j]+a[i]-a[i+1]); if(a[i]<a[j-1]) dp[i][j]=max(dp[i][j],dp[i][j-2]+a[j]-a[j-1]); if(a[i+1]<a[j]) dp[i][j]=max(dp[i][j],dp[i+1][j-1]+a[i]-a[j]); if(a[i]>=a[j-1]) dp[i][j]=max(dp[i][j],dp[i+1][j-1]+a[j]-a[i]); //dp[i][j]=max(max(dp[i+2][j]+a[i]-a[i+1],dp[i][j-2]+a[j]-a[j-1]), // max(dp[i+1][j-1]+a[i]-a[j],dp[i+1][j-1]+a[j]-a[i])); //dp[i+1][j-1]+a[i]-a[j] } /*for(i=1;i<=n;i++) { for(j=1;j<=n;j++) cout<<dp[i][j]<<" "; cout<<endl; }*/ t++; printf("In game %d, the greedy strategy might lose by as many as %d points./n",t,dp[1][n]); } return 0;}