hdu 4753 Fishhead’s Little Game （狀壓+記憶化搜尋）

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: Tom200Hint　　In case 1, Tom200 gets two points when she add edge 5 -> 6, two points in edge 6 -> 7, one point in 4 -> 8.

 
Source2013 ACM/ICPC Asia Regional Nanjing Online

#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)  // 尚未訪問這條邊        {            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;}