HDU 4285 circuits 第37届ACM/ICPC天津赛区网络赛（插头DP）

2023年2月8日 419次阅读来源: 算法小白

circuits

Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 418 Accepted Submission(s): 141

Problem Description 　　Given a map of N * M (2 <= N, M <= 12) , ‘.’ means empty, ‘*’ means walls. You need to build K circuits and no circuits could be nested in another. A circuit is a route connecting adjacent cells in a cell sequence, and also connect the first cell and the last cell. Each cell should be exactly in one circuit. How many ways do we have?

Input 　　The first line of input has an integer T, number of cases.

　　For each case:

　　The first line has three integers N M K, as described above.

　　Then the following N lines each has M characters, ‘.’ or ‘*’.

Output 　　For each case output one lines.

　　Each line is the answer % 1000000007 to the case.

Sample Input 2 4 4 1 **.. …. …. …. 4 4 1 …. …. …. ….

Sample Output 2 6

Source
2012 ACM/ICPC Asia Regional Tianjin Online

Recommend liuyiding 插头DP的题目。要找出刚好是K条回路的方案数。而且要避免形成环套环的情况。增加标记位来记录形成的回路数。不形成环的话就是在形成新的回路时，两边的插头个数要为偶数。此题很容易超时，环套环的没有避免又会WA。

/*
HDU 4285
要形成刚好K条回路的方法数
要避免环套环的情况。
所以形成回路时，要保证两边的插头数是偶数

G++ 11265ms  11820K
C++ 10656ms  11764K

*/

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

const int MAXD=15;
const int STATE=1000010;
const int HASH=300007;//这个大一点可以防止TLE,但是容易MLE
const int MOD=1000000007;

int N,M,K;
int maze[MAXD][MAXD];
int code[MAXD];
int ch[MAXD];
int num;//圈的个数
struct HASHMAP
{
    int head[HASH],next[STATE],size;
    long long state[STATE];
    int f[STATE];
    void init()
    {
        size=0;
        memset(head,-1,sizeof(head));
    }
    void push(long long st,int ans)
    {
        int i;
        int h=st%HASH;
        for(i=head[h];i!=-1;i=next[i])
          if(state[i]==st)
          {
              f[i]+=ans;
              f[i]%=MOD;
              return;
          }
        state[size]=st;
        f[size]=ans;
        next[size]=head[h];
        head[h]=size++;
    }
}hm[2];
void decode(int *code,int m,long long  st)
{
    num=st&63;
    st>>=6;
    for(int i=m;i>=0;i--)
    {
        code[i]=st&7;
        st>>=3;
    }
}
long long encode(int *code,int m)//最小表示法
{
    int cnt=1;
    memset(ch,-1,sizeof(ch));
    ch[0]=0;
    long long st=0;
    for(int i=0;i<=m;i++)
    {
        if(ch
]==-1)ch
]=cnt++;
 code[i]=ch
];
 st<<=3;
 st|=code[i];
 }
 st<<=6;
 st|=num;
 return st;
 }
 void shift(int *code,int m)
 {
 for(int i=m;i>0;i--)code[i]=code[i-1];
 code[0]=0;
 }
 void dpblank(int i,int j,int cur)
 {
 int k,left,up;
 for(k=0;k<hm[cur].size;k++)
 {
 decode(code,M,hm[cur].state[k]);
 left=code[j-1];
 up=code[j];
 if(left&&up)
 {
 if(left==up)
 {
 if(num>=K)continue;
 int t=0;
 //要避免环套环的情况，需要两边插头数为偶数
 for(int p=0;p<j-1;p++)
 if(code[p])t++;
 if(t&1)continue;
 if(num<K)
 {
 num++;
 code[j-1]=code[j]=0;
 hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k]);
 }
 }
 else
 {
 code[j-1]=code[j]=0;
 for(int t=0;t<=M;t++)
 if(code[t]==up)
 code[t]=left;
 hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k]);
 }
 }
 else if(left||up)
 {
 int t;
 if(left)t=left;
 else t=up;
 if(maze[i][j+1])
 {
 code[j-1]=0;
 code[j]=t;
 hm[cur^1].push(encode(code,M),hm[cur].f[k]);
 }
 if(maze[i+1][j])
 {
 code[j]=0;
 code[j-1]=t;
 hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k]);
 }
 }
 else
 {
 if(maze[i][j+1]&&maze[i+1][j])
 {
 code[j-1]=code[j]=13;
 hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k]);
 }
 }
 }
 }
 void dpblock(int i,int j,int cur)
 {
 int k;
 for(k=0;k<hm[cur].size;k++)
 {
 decode(code,M,hm[cur].state[k]);
 code[j-1]=code[j]=0;
 hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k]);
 }
 }
 char str[20];
 void init()
 {
 scanf("%d%d%d",&N,&M,&K);
 memset(maze,0,sizeof(maze));
 for(int i=1;i<=N;i++)
 {
 scanf("%s",&str);
 for(int j=1;j<=M;j++)
 if(str[j-1]=='.')
 maze[i][j]=1;
 }
 }
 void solve()
 {
 int i,j,cur=0;
 hm[cur].init();
 hm[cur].push(0,1);
 for(i=1;i<=N;i++)
 for(j=1;j<=M;j++)
 {
 hm[cur^1].init();
 if(maze[i][j])dpblank(i,j,cur);
 else dpblock(i,j,cur);
 cur^=1;
 }
 int ans=0;
 for(i=0;i<hm[cur].size;i++)
 if(hm[cur].state[i]==K)
 {
 ans+=hm[cur].f[i];
 ans%=MOD;
 }
 printf("%d\n",ans);
}
 int main()
 {
 //freopen("in.txt","r",stdin);
 //freopen("out.txt","w",stdout);
 int T;
 scanf("%d",&T);
 while(T--)
 {
 init();
 solve();
 }
 return 0;
 }
/*
 Sample Input
 2
 4 4 1
 **..
 ....
 ....
 ....
 4 4 1
 ....
 ....
 ....
 ....
Sample Output
 2
 6
*/
 
 
另外一个记录方法：
/*
HDU  4285
要形成刚好K条回路的方法数
要避免环套环的情况。
所以形成回路时，要保证两边的插头数是偶数
G++ 11765ms  12560K
C++  11656ms 12504K
*/

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

const int MAXD=15;
const int STATE=1000010;
const int HASH=100007;
const int MOD=1000000007;

int N,M,K;
int maze[MAXD][MAXD];
int code[MAXD];
int ch[MAXD];

struct HASHMAP
{
    int head[HASH],next[STATE],size;
    long long state[STATE];
    int f[STATE];
    int cir[STATE];//形成的圈的个数
    void init()
    {
        size=0;
        memset(head,-1,sizeof(head));
    }
    void push(long long st,int ans,int _cir)
    {
        int i,h=st%HASH;
        for(i=head[h];i!=-1;i=next[i])
          if(state[i]==st&&cir[i]==_cir)
          {
              f[i]+=ans;
              f[i]%=MOD;
              return;
          }
        state[size]=st;
        f[size]=ans;
        cir[size]=_cir;
        next[size]=head[h];
        head[h]=size++;
    }
}hm[2];
void decode(int *code,int m,long long  st)
{
    for(int i=m;i>=0;i--)
    {
        code[i]=st&7;
        st>>=3;
    }
}
long long encode(int *code,int m)//最小表示法
{
    int cnt=1;
    memset(ch,-1,sizeof(ch));
    ch[0]=0;
    long long st=0;
    for(int i=0;i<=m;i++)
    {
        if(ch
]==-1)ch
]=cnt++;
 code[i]=ch
];
 st<<=3;
 st|=code[i];
 }
 return st;
 }
 void shift(int *code,int m)
 {
 for(int i=m;i>0;i--)code[i]=code[i-1];
 code[0]=0;
 }
 void dpblank(int i,int j,int cur)
 {
 int k,left,up;
 for(k=0;k<hm[cur].size;k++)
 {
 decode(code,M,hm[cur].state[k]);
 left=code[j-1];
 up=code[j];
 if(left&&up)
 {
 if(left==up)
 {
 if(hm[cur].cir[k]>=K)continue;
 int t=0;
 for(int p=0;p<j-1;p++)
 if(code[p])t++;
 if(t&1)continue;
 if(hm[cur].cir[k]<K)
 {
 code[j-1]=code[j]=0;
 hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k],hm[cur].cir[k]+1);
 }
 }
 else
 {
 code[j-1]=code[j]=0;
 for(int t=0;t<=M;t++)
 if(code[t]==up)
 code[t]=left;
 hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k],hm[cur].cir[k]);
 }
 }
 else if(left||up)
 {
 int t;
 if(left)t=left;
 else t=up;
 if(maze[i][j+1])
 {
 code[j-1]=0;
 code[j]=t;
 hm[cur^1].push(encode(code,M),hm[cur].f[k],hm[cur].cir[k]);
 }
 if(maze[i+1][j])
 {
 code[j]=0;
 code[j-1]=t;
 hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k],hm[cur].cir[k]);
 }
 }
 else
 {
 if(maze[i][j+1]&&maze[i+1][j])
 {
 code[j-1]=code[j]=13;
 hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k],hm[cur].cir[k]);
 }
 }
 }
 }
 void dpblock(int i,int j,int cur)
 {
 int k;
 for(k=0;k<hm[cur].size;k++)
 {
 decode(code,M,hm[cur].state[k]);
 code[j-1]=code[j]=0;
 hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k],hm[cur].cir[k]);
 }
 }
 char str[20];
 void init()
 {
 scanf("%d%d%d",&N,&M,&K);
 memset(maze,0,sizeof(maze));
 for(int i=1;i<=N;i++)
 {
 scanf("%s",&str);
 for(int j=1;j<=M;j++)
 if(str[j-1]=='.')
 maze[i][j]=1;
 }
 }
 void solve()
 {
 int i,j,cur=0;
 hm[cur].init();
 hm[cur].push(0,1,0);
 for(i=1;i<=N;i++)
 for(j=1;j<=M;j++)
 {
 hm[cur^1].init();
 if(maze[i][j])dpblank(i,j,cur);
 else dpblock(i,j,cur);
 cur^=1;
 }
 int ans=0;
 for(i=0;i<hm[cur].size;i++)
 if(hm[cur].cir[i]==K)
 {
 ans+=hm[cur].f[i];
 ans%=MOD;
 }
 printf("%d\n",ans);
}
 int main()
 {
 // freopen("in.txt","r",stdin);
 // freopen("out.txt","w",stdout);
 int T;
 scanf("%d",&T);
 while(T--)
 {
 init();
 solve();
 }
 return 0;
 }
/*
 Sample Input
 2
 4 4 1
 **..
 ....
 ....
 ....
 4 4 1
 ....
 ....
 ....
 ....
Sample Output
 2
 6
*/

    原文作者：算法小白
    原文地址: https://www.cnblogs.com/kuangbin/archive/2012/10/02/2710308.html
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