Triangle War
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 2685   Accepted: 1061

Description

Triangle War is a two-player game played on the following triangular grid: 


Two players, A and B, take turns filling in any dotted line connecting two dots, with A starting first. Once a line is filled, it cannot be filled again. If the line filled by a player completes one or more triangles, she owns the completed triangles and she
is awarded another turn (i.e. the opponent skips a turn). The game ends after all dotted lines are filled in, and the player with the most triangles wins the game. The difference in the number of triangles owned by the two players is not important. 

For example, if A fills in the line between 2 and 5 in the partial game on the left below: 


Then, she owns the triangle labelled A and takes another turn to fill in the line between 3 and 5. B can now own 3 triangles (if he wishes) by filling in the line between 2 and 3, then the one between 5 and 6, and finally the one between 6 and 9. B would then
make one more move before it is A's turn again. 
In this problem, you are given a number of moves that have already been made. From the partial game, you should determine which player will win assuming that each player plays a perfect game from that point on. That is, assume that each player always chooses
the play that leads to the best possible outcome for himself/herself.

Input

You will be given a number of games in the input. The first line of input is a positive integer indicating the number of games to follow. Each game starts with an integer 6 <= m <= 18 indicating the number of moves that have been made in the game. The next
m lines indicate the moves made by the two players in order, each of the form i j (with i < j) indicating that the line between i and j is filled in that move. You may assume that all given moves are legal.

Output

For each game, print the game number and the result on one line as shown below. If A wins, print the sentence "A wins." If B wins, print "B wins."

Sample Input

4
6
2 4
4 5
5 9
3 6
2 5
3 5
7
2 4
4 5
5 9
3 6
2 5
3 5
7 8
6
1 2
2 3
1 3
2 4
2 5
4 5
10
1 2
2 5
3 6
5 8
4 7
6 10
2 4
4 5
4 8
7 8

Sample Output

Game 1: B wins.
Game 2: A wins.
Game 3: A wins.
Game 4: B wins.

Source

题意:

两个人玩游戏,依次在三角形上放边,假设能构成三角形。则奖励继续该此人放,问最后得到的三角形多。

思路:

给边编号,记忆化搜索即可。做过好多这样的题。就不多写思路了。

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 50005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std; int n,m,ans,cnt,tot,flag;
bool vis[10];
int dp[300000],mp[15][15],sc[5];
int tri[9][3]=
{
0,1,2,3,4,7,2,4,5,5,6,8,9,10,15,
7,10,11,11,12,16,8,12,13,13,14,17
}; int cal(int s)
{
int i,j,t=0;
for(j=0; j<9; j++)
{
if((s&(1<<tri[j][0]))&&(s&(1<<tri[j][1]))&&(s&(1<<tri[j][2]))) t++;
}
return t;
}
int dfs(int state,int score)
{
if(dp[state]!=-1) return dp[state];
int i,j,t,tst,num,best=0,tmp;
num=9-score;
for(i=0; i<=17; i++)
{
if(state&(1<<i)) continue ;
tst=state|(1<<i);
t=cal(tst);
if(t>num)
{
tmp=t-num+dfs(tst,score-(t-num));
best=max(best,tmp);
}
else
{
tmp=score-dfs(tst,score);
best=max(best,tmp);
}
}
dp[state]=best;
return best;
}
int main()
{
int i,j,t,test=0;
mp[1][2]=0;mp[1][3]=1;mp[2][3]=2;mp[2][4]=3;mp[2][5]=4;mp[3][5]=5;
mp[3][6]=6;mp[4][5]=7;mp[5][6]=8;mp[4][7]=9;mp[4][8]=10;mp[5][8]=11;
mp[5][9]=12;mp[6][9]=13;mp[6][10]=14;mp[7][8]=15;mp[8][9]=16;mp[9][10]=17;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
tot=0;
int x,y,z,turn=0,num=0;
sc[0]=sc[1]=0;
memset(vis,0,sizeof(vis));
for(i=1; i<=n; i++)
{
scanf("%d%d",&x,&y);
z=mp[x][y];
tot|=(1<<z);
flag=0;
for(j=0; j<9; j++)
{
if(vis[j]) continue ;
if((tot&(1<<tri[j][0]))&&(tot&(1<<tri[j][1]))&&(tot&(1<<tri[j][2])))
{
vis[j]=1;
num++;
flag=1;
sc[turn]++;
}
}
if(!flag) turn^=1;
}
memset(dp,-1,sizeof(dp));
z=dfs(tot,9-num);
sc[turn]+=z;
sc[turn^1]+=(9-num-z);
if(sc[0]>sc[1]) printf("Game %d: A wins.\n",++test);
else printf("Game %d: B wins.\n",++test);
}
return 0;
}

poj 1085 Triangle War (状压+记忆化搜索)的更多相关文章

  1. Luogu P2831 愤怒的小鸟(状压+记忆化搜索)

    P2831 愤怒的小鸟 题意 题目描述 Kiana最近沉迷于一款神奇的游戏无法自拔. 简单来说,这款游戏是在一个平面上进行的. 有一架弹弓位于\((0,0)\)处,每次Kiana可以用它向第一象限发射 ...

  2. poj 1085 Triangle War 博弈论+记忆化搜索

    思路:总共有18条边,9个三角形. 极大极小化搜索+剪枝比较慢,所以用记忆化搜索!! 用state存放当前的加边后的状态,并判断是否构成三角形,找出最优解. 代码如下: #include<ios ...

  3. POJ 1191 棋盘分割 【DFS记忆化搜索经典】

    题目传送门:http://poj.org/problem?id=1191 棋盘分割 Time Limit: 1000MS   Memory Limit: 10000K Total Submission ...

  4. POJ 1579 Function Run Fun 【记忆化搜索入门】

    题目传送门:http://poj.org/problem?id=1579 Function Run Fun Time Limit: 1000MS   Memory Limit: 10000K Tota ...

  5. (中等) POJ 1054 The Troublesome Frog,记忆化搜索。

    Description In Korea, the naughtiness of the cheonggaeguri, a small frog, is legendary. This is a we ...

  6. POJ 3249 Test for Job (记忆化搜索)

    Test for Job Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 11830   Accepted: 2814 Des ...

  7. HDU 2517 / POJ 1191 棋盘分割 区间DP / 记忆化搜索

    题目链接: 黑书 P116 HDU 2157 棋盘分割 POJ 1191 棋盘分割 分析:  枚举所有可能的切割方法. 但如果用递归的方法要加上记忆搜索, 不能会超时... 代码: #include& ...

  8. poj 1088 滑雪(区间dp+记忆化搜索)

    题目链接:http://poj.org/problem?id=1088 思路分析: 1>状态定义:状态dp[i][j]表示在位置map[i][j]可以滑雪的最长区域长度: 2>状态转移方程 ...

  9. POJ 1088: 滑雪(经典 DP+记忆化搜索)

    滑雪 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 74996   Accepted: 27818 Description ...

随机推荐

  1. SQL Server分页查询进化史

    分页查询一直SQL Server的一个硬伤,就是是经过一些进化,比起MySql的limit还是有一些差距. 一.条件过滤(适应用所有版本) 条件过滤的方法有很多,而思路就是利用集合的差集选择出目标集合 ...

  2. Django项目:CRM(客户关系管理系统)--27--19PerfectCRM实现King_admin数据修改

    登陆密码设置参考 http://www.cnblogs.com/ujq3/p/8553784.html {#table_data_list.html#} {## ————————08PerfectCR ...

  3. LUOGU P2939 [USACO09FEB]改造路Revamping Trails

    题意翻译 约翰一共有N)个牧场.由M条布满尘埃的小径连接.小径可 以双向通行.每天早上约翰从牧场1出发到牧场N去给奶牛检查身体. 通过每条小径都需要消耗一定的时间.约翰打算升级其中K条小径,使之成为高 ...

  4. Excel函数学习:HLOOKUP函数

    Excel函数学习:HLOOKUP函数 HLOOKUP函数查找表的第一行中的值,返回该表中与找到的值在同一列的另一个值. 什么情况下使用HLOOKUP? HLOOKUP函数可以在查找行中找到精确匹配值 ...

  5. C# Action 和Func

    https://www.cnblogs.com/LipeiNet/p/4694225.html

  6. Scanner读取记事本文件内容为空的解决办法

    原因:记事本txt文件中含有中文,windows记事本编码方式为gbk,但是eclipse中为utf-8,所以需要在Scanner中指定编码方式.

  7. solr问题missing content stream

    在使用solrj建立索引的时候,报错:missing content stream; 原因在于 HttpSolrServer httpSolrServer = new HttpSolrServer(s ...

  8. 洛谷P2178 [NOI2015]品酒大会

    题目描述 一年一度的“幻影阁夏日品酒大会”隆重开幕了.大会包含品尝和趣味挑战 两个环节,分别向优胜者颁发“首席品酒家”和“首席猎手”两个奖项,吸引了众多品酒师参加. 在大会的晚餐上,调酒师 Rainb ...

  9. MySQL Daemon failed to start错误解决办法是什么呢?

    首先我尝试用命令:service mysql start 来启动服务,但是提示: MySQL Daemon failed to start 一开始出现这个问题我很方,然后开始查,说什么的都有,然后看到 ...

  10. angular.module()创建、获取、注册angular中的模块

    // 传递参数不止一个,代表新建模块;空数组代表该模块不依赖其他模块 var createModule = angular.module("myModule", []); // 只 ...