周六第一次做usaco玩，bronze的輕松切掉，然后申請promote，下午批準，話說rob 效率好高啊~ 于是繼續做silver 就遇到這個題- -！糾結了半天放棄····知道是dp 也考慮了方法就是 理不清楚；不知道是不是一天沒吃飯的緣故·····

今天題解出來了~ 先看了大概思路 然后自己寫出來了~

題目：

Farmer John's cows like to play coin games so FJ has invented with
a new two-player coin game called Xoinc for them.

Initially a stack of N (5 <= N <= 2,000) coins sits on the ground;
coin i from the top has integer value C_i (1 <= C_i <= 100,000).
The first player starts the game by taking the top one or two coins
(C_1 and maybe C_2) from the stack. If the first player takes just
the top coin, the second player may take the following one or two
coins in the next turn. If the first player takes two coins then
the second player may take the top one, two, three or four coins
from the stack. In each turn, the current player must take at least
one coin and at most two times the amount of coins last taken by
the opposing player. The game is over when there are no more coins
to take.

Afterwards, they can use the value of the coins they have taken
from the stack to buy treats from FJ, so naturally, their purpose
in the game is to maximize the total value of the coins they take.
Assuming the second player plays optimally to maximize his own
winnings, what is the highest total value that the first player can
have when the game is over?

MEMORY LIMIT: 20 MB

PROBLEM NAME: xoinc

INPUT FORMAT:

* Line 1: A single integer: N

* Lines 2..N+1: Line i+1 contains a single integer: C_i

SAMPLE INPUT (file xoinc.in):

5
1
3
1
7
2

簡單來說就是兩個人輪流取coins，每個人每次取得個數為1- 2*n;n為上一輪對方取得數目，

求兩個人都是用最佳策略，先取得那個家伙最多能拿到多少硬幣。貌似可以算是簡單博弈論的思想

思路：

        coins[1···N] 從下到上 sum[1···N] 剩下 i個的和

        找到無后效性的子問題?？紤]在還剩下p個錢幣時候的情況，此時可以拿k個錢

由于條件，k的大小受上一輪拿的個數i的限制 ，所以我們要加上一個變量i。得到

dp[p][i]這個子問題。那么容易得到

dp[p][i]=max(1=<k<=i*2){SuM(p to p-k+1)+SuM(p-k to 1)-dp[p-k][k]}

            =max(1=<k<=i*2){sum[p]-dp[p-k][k]}

按照這個可以得到一個O（N^3）的算法

oidsolve(){
   for(inti=1;i<=N;i++)//剩下i個
        for(intj=1;j<=N;j++)//上一人拿了j 個
            for(intk=1;k<=j*2&&i-k>=0;k++){
                dp[i][j]=max(dp[i][j],sum[1]-sum[i+1]-dp[i-k][k]);
            }
    ret=dp[N][1];
}

 三重遞歸 ，最多可以過500的數據量  觀察可以得出 dp[p][j] 和 dp[p][j+1] 的計算有很多的重疊
因為 上次拿了j+1 則可以比 dp[p][j] 多拿 2 個 

然后，由于考慮j的范圍 應該為 N-i+1

這樣得到了最終代碼：

scanf("%d",&N);
for(int i=1;i<=N;i++)    scanf("%d",coins+i);//{fin>>coins[i]; }
sum[0]=0;
for(int i=1;i<=N;i++)     sum[i]=sum[i-1]+coins[N-i+1]; 
for(int i=1;i<=N;i++)        //剩下i個
for(int j=1;j<= N-i +1;j++){ // 上次拿了j個
if(dp[i][j]<dp[i][j-1])dp[i][j]=dp[i][j-1];
if(2*j-1<=i&&dp[i][j]<sum[i]-dp[i-2*j+1][2*j-1]) dp[i][j]=sum[i]-dp[i-2*j+1][2*j-1];
if(2*j<=i&&dp[i][j]<sum[i]-dp[i-2*j][2*j]) dp[i][j]= sum[i]-dp[i-2*j][2*j];
}
printf("%d\n",dp[N][1]);

很晚了 ，先寫這么多 ，有空把bronze的寫了

3個月后注：事實證明，當時么有時間 ~以后更沒有時間 ~~~ hoho`````````~~~~~~~``````````