亂八七糟的代碼

周六第一次做usaco玩，bronze的輕松切掉，然后申請(qǐng)promote，下午批準(zhǔn)，話說(shuō)rob 效率好高啊~ 于是繼續(xù)做silver 就遇到這個(gè)題- -！糾結(jié)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(méi)吃飯的緣故·····

今天題解出來(lái)了~ 先看了大概思路 然后自己寫(xiě)出來(lái)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(jiǎn)單來(lái)說(shuō)就是兩個(gè)人輪流取coins，每個(gè)人每次取得個(gè)數(shù)為1- 2*n;n為上一輪對(duì)方取得數(shù)目，

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

思路：

        coins[1···N] 從下到上 sum[1···N] 剩下 i個(gè)的和

        找到無(wú)后效性的子問(wèn)題。考慮在還剩下p個(gè)錢(qián)幣時(shí)候的情況，此時(shí)可以拿k個(gè)錢(qián)

由于條件，k的大小受上一輪拿的個(gè)數(shù)i的限制 ，所以我們要加上一個(gè)變量i。得到

dp[p][i]這個(gè)子問(wèn)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(gè)可以得到一個(gè)O（N^3）的算法

oidsolve(){
   for(inti=1;i<=N;i++)//剩下i個(gè)
        for(intj=1;j<=N;j++)//上一人拿了j 個(gè)
            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(guò)500的數(shù)據(jù)量  觀察可以得出 dp[p][j] 和 dp[p][j+1] 的計(jì)算有很多的重疊
因?yàn)?上次拿了j+1 則可以比 dp[p][j] 多拿 2 個(gè) 

然后，由于考慮j的范圍 應(yīng)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(gè)
for(int j=1;j<= N-i +1;j++){ // 上次拿了j個(gè)
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(xiě)這么多 ，有空把bronze的寫(xiě)了

3個(gè)月后注：事實(shí)證明，當(dāng)時(shí)么有時(shí)間 ~以后更沒(méi)有時(shí)間 ~~~ hoho`````````~~~~~~~``````````