hdu 4064 Carcassonne

最新推荐文章于 2022-02-25 19:38:17 发布

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本文深入探讨了游戏开发领域的关键技术，包括游戏引擎、编程语言、硬件优化等，并重点阐述了AI音视频处理的应用场景和实现方法，如语义识别、物体检测、语音变声等。通过实例分析，揭示了技术融合带来的创新解决方案。

Problem Description

Carcassonne is a tile-based board game for two to five players.
Square tiles are printed by city segments,road segments and field segments.

The rule of the game is to put the tiles alternately. Two tiles share one edge should exactly connect to each other, that is, city segments should be linked to city segments, road to road, and field to field.

To simplify the problem, we only consider putting tiles:
Given n*m tiles. You can rotate each tile, but not flip top to bottom, and not change their order.
How many ways could you rotate them to make them follow the rules mentioned above?

Input

The first line is a number T(1<=T<=50), represents the number of case. The next T blocks follow each indicates a case.
Each case starts with two number N,M(0<N,M<=12)
Then N*M lines follow,each line contains M four-character clockwise.
'C' indicate City.
'R' indicate Road.
'F' indicate Field.

Output

For each case, output the number of ways mod 1,000,000,007.(as shown in the sample output)

Sample Input

  
   3
1 1
RRRR
1 2
RRRF FCCC
8 8
FCFF RRFC FRCR FRFR RCCR FFCC RRFF CRFR
FRRC FRFR CCCR FCFC CRRC CRRR FRCR FRFR
RRCR FRRR CCCR FFFC RRFF RFCR CCFF FCCC
CFCF RRFF CRFR FFRR FRRF CCRR FFFC CRRF
CFRR FFFF FFFF RRFF RRRR RCRR FFCC RFRF
RRCF FRFR FRRR FRFR RCCR RCCC CFFC RFRF
CFCF FRFF RRFF FFFF CFFF CFFF FRFF RFRR
CCRR FCFC FCCC FCCC FFCC FCCF FFCC RFRF

Sample Output

  
   Case 1: 4
Case 2: 1
Case 3: 1048576

Source

The 36th ACM/ICPC Asia Regional Fuzhou Site —— Online Contest

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代码

#include <stdio.h>
#include <string.h>

#define MOD 1000000007

int map[15][15][5];
char str[5];
int n,m;
long long dp[2][600000];
int three[15];
bool flag;
int tag;


void DFS(int h,int t,int shang,int xia,int right)
{
    int i,j,n,ss,xx,zz,yy;
    if (t==m)
    {
      //  printf("%d..\n",tag);
        if (h==0)
        {
            dp[flag][xia]+=tag;
        }
        else
        {
            dp[flag][xia]=(dp[flag][xia]+dp[flag^1][shang]*tag)%MOD;
        }
        return;
    }
    zz=map[h][t][0];
    ss=map[h][t][1];
    yy=map[h][t][2];
    xx=map[h][t][3];
    if (zz==ss && zz==yy && zz==xx && ss==yy && ss==xx && yy==xx)
    {
        tag*=4;
        if (zz==right || right==-1) DFS(h,t+1,shang*3+ss,xia*3+xx,yy);
        tag/=4;
        return;
    }
    for (i=0;i<4;i++)
    {
        zz=map[h][t][(0+i)%4];
        ss=map[h][t][(1+i)%4];
        yy=map[h][t][(2+i)%4];
        xx=map[h][t][(3+i)%4];
        if (zz==right || right==-1) DFS(h,t+1,shang*3+ss,xia*3+xx,yy);
    }
}

int Change(char s)
{
    if (s=='C') return 0;
    if (s=='R') return 1;
    return 2;
}


int main()
{
    int i,j,T,k,kk,ans,cnt;
    cnt=1;
    three[0]=1;
    for (i=1;i<=13;i++)
    {
        three[i]=three[i-1]*3;
    }
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for (i=0;i<n;i++)
        {
            for (j=0;j<m;j++)
            {
                scanf("%s",str);
                for (k=0;k<4;k++)
                {
                    map[i][j][k]=Change(str[k]);
                }
            }
        }
        memset(dp,0,sizeof(dp));
        flag=0;
        for (i=0;i<n;i++)
        {
            tag=1;
            DFS(i,0,0,0,-1);
            flag=1^flag;
            for (j=0;j<three[m];j++) dp[flag][j]=0;
        }
        ans=0;
        for (i=0;i<three[m];i++)
        {
            ans=(ans+dp[1^flag][i])%MOD;
        }
        printf("Case %d: %d\n",cnt++,ans);
    }
    return 0;
}