新學(xué)習(xí)了知識點；

博弈樹 alpha-beta 剪枝：

https://blog.csdn.net/qq_27008079/article/details/60869054

還有一個國外的網(wǎng)站， 講的很明晰：

http://web.cs.ucla.edu/~rosen/161/notes/alphabeta.html

Rake It In

時間限制:?1 Sec?? 內(nèi)存限制:?128 MB
提交:?73?? 解決:?13
[ 提交][ 狀態(tài)][ 討論版][命題人: admin]

題目描述

The designers have come up with a new simple game called “Rake It In”. Two players, Alice and Bob, initially? select an integer k and initialize a score indicator. An 4 × 4 board is created with 16 values placed on the board.
Starting with player Alice, each player in a round selects a 2 × 2 region of the board, adding the sum of values? in the region to the score indicator, and then rotating these four values 90 degrees counterclockwise.
After 2k rounds in total, each player has made decision in k times. The ultimate goal of Alice is to maximize? the ?nal score. However for Bob, his goal is to minimize the ?nal score.
In order to test how good this game is, you are hired to write a program which can play the game. Speci?cally,? given the starting con?guration, they would like a program to determine the ?nal score when both players are? entirely rational.

輸入

The input contains several test cases and the first line provides an integer t (1 ≤ t ≤ 200) which is the number of? test cases.
Each case contains five lines. The first line provides the integer k (1 ≤ k ≤ 3). Each of the following four lines? contains four integers indicating the values on the board initially. All values are integers between 1 to 10.

輸出

For each case, output an integer in a line which is the predicted ?nal score.

樣例輸入

4 1 1 1 2 2 1 1 2 2 3 3 4 4 3 3 4 4 2 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 3 1 1 4 4 4 4 1 1 1 1 4 4 1 4 1 4 3 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1

樣例輸出

20 40 63 71

提示

來源

ICPC2017?Naning?

【思路】

? ?用alph-beta 剪枝， 不然的話會超時， 說白了 就是 DFS

【code】

/* * Date:4/8/2018 * Tile: ACM ICPC nanning * Category: 博弈樹 alpha-beta 剪枝 (DFS) */ #include <iostream> #include <bits/stdc++.h>typedef long long ll; const int MAXN=1e5+5; const int INF=0x3f3f3f3f;using namespace std;int k; int mp[5][5]; int dfs(int h,int a[5][5],int x,int y,int player,int alpha,int beta,int s) {int sum=0;int newa[5][5];for(int i=1;i<=4;i++)for(int j=1;j<=4;j++)newa[i][j]=a[i][j];if(h){for(int i=x;i<=x+1;i++)for(int j=y;j<=y+1;j++)sum+=newa[i][j];swap(newa[x][y],newa[x+1][y]);swap(newa[x][y],newa[x+1][y+1]);swap(newa[x][y],newa[x][y+1]);}if( h==2*k)return s+sum;for(int i=1;i<=3;i++){for(int j=1;j<=3;j++){if(player)beta=min(beta,dfs(h+1,newa,i,j,player^1,alpha,beta,s+sum));elsealpha=max(alpha,dfs(h+1,newa,i,j,player^1,alpha,beta,s+sum));if(beta<=alpha){break;}}if(beta<=alpha){break;}}return player?beta:alpha; } int main() {int t;scanf("%d",&t);while(t--){scanf("%d",&k);for(int i=1;i<=4;i++)for(int j=1;j<=4;j++)scanf("%d",&mp[i][j]);int ans=dfs(0,mp,0,0,0,-INF,INF,0);printf("%d\n",ans);}return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/sizaif/p/8822640.html

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