hdu 4753 Fishhead’s Little Game （状压+记忆化搜索）

Fishhead’s Little Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 517    Accepted Submission(s): 124

Problem Description

　　　There is a 3 by 3 grid and each vertex is assigned a number.


　　　It looks like JiuGongGe, but they are different, for we are not going to fill the cell but the edge. For instance,


adding edge 6 –> 10
　　　The rule of this game is that each player takes turns to add an edge. You will get one point if the edge you just added, together with edges already added before, forms a new square (only square of size 1 is considered). Of course, you get two points if that edge forms two squares. Notice that an edge can be added only once.


forming two squares to get two points
　　Tom200 and Jerry404 is playing this little game, and have played n rounds when Fishhead comes in. Fishhead wants to know who will be the winner. Can you help him? Assume that Tom200 and Jerry404 are clever enough to make optimal decisions in each round. Every Game starts from Tom200.

 

Input

　　The first line of the input contains a single integer T (T <= 100), the number of test cases.
　　For each case, the first line contains an integers n (12 <= n <= 24), which means they have taken total n rounds in turn. Next n lines each contains two integers a, b (a, b <= 16) representing the two endpoints of the edge.

 

Output

　　For each case output one line "Case #X: ", representing the Xth case, starting from 1. If Tom200 wins, print "Tom200" on one line, print "Jerry404" otherwise.

 

Sample Input

1151 21 52 65 96 109 105 62 33 77 1110 113 46 77 84 8

 

Sample Output

Case #1: Tom200
Hint
　　In case 1, Tom200 gets two points when she add edge 5 -> 6, two points in edge 6 -> 7, one point in 4 -> 8.
 

 

Source

2013 ACM/ICPC Asia Regional Nanjing Online

题意：

在3*3的方格中，有4*4=16个点，标号分别为1~16，A、B两人轮流玩游戏，每次可以添加一条边（相邻节点），如果恰好能够凑成一个边长为1的正方形则得一分，两个的话得2分。给定两人放的n边后，求最终格局谁会胜。

思路：

状压+记忆化搜索。

将没有访问的边记录下来，用一个数表示这些边的状态实现状压。假设dfs(state)能生成一个该状态下的最优解，用dp[]保存下来就实现记忆化了。

一个state还剩sum分可加，对这个state进行加边（将其中一位 置为1），加这条边后能的 t 分，则

score=sum-dfs(newstate,sum-t)；

取每种情况的score的最大值就够了。

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 30#define mod 1000000000#define INF 0x3f3f3f3fusing namespace std;int n,m,ans,cnt;int x[maxn],dp[10000];bool vis[maxn];int mp[maxn][maxn];int s[9][4]={    1,4,5,8,    2,5,6,9,    3,6,7,10,    8,11,12,15,    9,12,13,16,    10,13,14,17,    15,18,19,22,    16,19,20,23,    17,20,21,24};int solve(){    int i,j,t=0;    for(i=0;i<9;i++)    {        if(vis[s[i][0]]&&vis[s[i][1]]&&vis[s[i][2]]&&vis[s[i][3]]) t++;    }    return t;}int isok(int ste,int k){    int i,j,u,s1,s2;    int vv[maxn]={0};    for(i=0;i<=cnt;i++)    {        u=1<<i;        if(ste&u) vv[x[i]]=1;    }    s1=s2=0;    for(i=0;i<9;i++)    {        if((vis[s[i][0]]||vv[s[i][0]])&&(vis[s[i][1]]||vv[s[i][1]])&&(vis[s[i][2]]||vv[s[i][2]])&&(vis[s[i][3]]||vv[s[i][3]])) s1++;    }    vv[x[k]]=1;    for(i=0;i<9;i++)    {        if((vis[s[i][0]]||vv[s[i][0]])&&(vis[s[i][1]]||vv[s[i][1]])&&(vis[s[i][2]]||vv[s[i][2]])&&(vis[s[i][3]]||vv[s[i][3]])) s2++;    }    return s2-s1;}int dfs(int ste,int sum){    if(dp[ste]!=-1) return dp[ste];    int i,j,st,t,tmp,best=0;    for(i=0;i<=cnt;i++)    {        st=1<<i;        if((st&ste)==0)  // 尚未访问这条边        {            st=st|ste;            tmp=isok(ste,i);            t=dfs(st,sum-tmp);            if(best<sum-t) best=sum-t;        }    }    dp[ste]=best;    return best;}int main(){    int i,j,t,u,v,s1,s2,T,J,test=0;    memset(mp,0,sizeof(mp));    mp[1][2]=1,mp[2][3]=2,mp[3][4]=3;    mp[1][5]=4,mp[2][6]=5,mp[3][7]=6,mp[4][8]=7;    mp[5][6]=8,mp[6][7]=9,mp[7][8]=10;    mp[5][9]=11,mp[6][10]=12,mp[7][11]=13,mp[8][12]=14;    mp[9][10]=15,mp[10][11]=16,mp[11][12]=17;    mp[9][13]=18,mp[10][14]=19,mp[11][15]=20,mp[12][16]=21;    mp[13][14]=22,mp[14][15]=23,mp[15][16]=24;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        memset(vis,0,sizeof(vis));        T=J=0;        s1=0;        for(i=1;i<=n;i++)        {            scanf("%d%d",&u,&v);            vis[mp[u][v]]=vis[mp[v][u]]=1;            s2=solve();            if(i&1) T+=s2-s1;            else J+=s2-s1;            s1=s2;        }        cnt=-1;        for(i=1;i<=24;i++)        {            if(!vis[i]) x[++cnt]=i;        }        m=0;        for(i=0;i<9;i++)        {            if(!vis[s[i][0]]||!vis[s[i][1]]||!vis[s[i][2]]||!vis[s[i][3]]) m++;        }        memset(dp,-1,sizeof(dp));        u=dfs(0,m);        printf("Case #%d: ",++test);        if(n&1)        {            if(T+m-u>J+u) printf("Tom200\n");            else printf("Jerry404\n");        }        else        {            if(T+u>J+m-u) printf("Tom200\n");            else printf("Jerry404\n");        }    }    return 0;}