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  • bzoj 1814 Ural 1519 Formula 1 插头DP

    1814: Ural 1519 Formula 1

    Time Limit: 1 Sec  Memory Limit: 64 MB
    Submit: 942  Solved: 356
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    Description

    Regardless of the fact, that Vologda could not get rights to hold the Winter Olympic games of 20**, it is well-known, that the city will conduct one of the Formula 1 events. Surely, for such an important thing a new race circuit should be built as well as hotels, restaurants, international airport - everything for Formula 1 fans, who will flood the city soon. But when all the hotels and a half of the restaurants were built, it appeared, that at the site for the future circuit a lot of gophers lived in their holes. Since we like animals very much, ecologists will never allow to build the race circuit over the holes. So now the mayor is sitting sadly in his office and looking at the map of the circuit with all the holes plotted on it. Problem Who will be smart enough to draw a plan of the circuit and keep the city from inevitable disgrace? Of course, only true professionals - battle-hardened programmers from the first team of local technical university!.. But our heroes were not looking for easy life and set much more difficult problem: "Certainly, our mayor will be glad, if we find how many ways of building the circuit are there!" - they said. It should be said, that the circuit in Vologda is going to be rather simple. It will be a rectangle N*M cells in size with a single circuit segment built through each cell. Each segment should be parallel to one of rectangle's sides, so only right-angled bends may be on the circuit. At the picture below two samples are given for N = M = 4 (gray squares mean gopher holes, and the bold black line means the race circuit). There are no other ways to build the circuit here. 一个 m * n 的棋盘,有的格子存在障碍,求经过所有非障碍格子的哈密顿回路个数

    Input

    The first line contains the integer numbers N and M (2 ≤ N, M ≤ 12). Each of the next N lines contains M characters, which are the corresponding cells of the rectangle. Character "." (full stop) means a cell, where a segment of the race circuit should be built, and character "*" (asterisk) - a cell, where a gopher hole is located.

    Output

    You should output the desired number of ways. It is guaranteed, that it does not exceed 2^63-1.

    Sample Input

    4 4
    **..
    ....
    ....
    ....



    Sample Output


    2

    HINT

    Source

     题解:什么叫做哈密顿回路,就是从一个点出发,经过所有点一次仅一次,

    所有可以知道路不能交叉。

    具体还是看cdqhttps://wenku.baidu.com/view/9cfbb16e011ca300a6c390d5.html

      1 #pragma GCC optimize(2)
      2 #pragma G++ optimize(2)
      3 #include<iostream>
      4 #include<algorithm>
      5 #include<cmath>
      6 #include<cstdio>
      7 #include<cstring>
      8 
      9 #define sz 200005
     10 #define ll long long
     11 using namespace std;
     12 
     13 int hs[sz],tot[2],sta[2][sz],bit[30];
     14 int n,m,c,ex,ey,mp[15][15];
     15 ll dp[2][sz],ans;
     16 char s[15];
     17 
     18 void add(int s,ll num)
     19 {
     20     int pos=s%sz;
     21     while (hs[pos]!=-1)
     22     {
     23         if (sta[c][hs[pos]]==s)
     24         {
     25             dp[c][hs[pos]]+=num;
     26             return;
     27         }
     28         pos++;
     29         if (pos==sz) pos=0;
     30     }
     31     dp[c][++tot[c]]=num;
     32     hs[pos]=tot[c];
     33     sta[c][tot[c]]=s;
     34 }
     35 void DP()
     36 {
     37     for (int i=1;i<=n;i++)
     38     {
     39         for (int k=1;k<=tot[c];k++)
     40             sta[c][k]<<=2;
     41         for (int j=1;j<=m;j++)
     42         {
     43             c^=1; tot[c]=0;
     44             memset(hs,-1,sizeof(hs));
     45             for (int k=1;k<=tot[c^1];k++)
     46             {
     47                 int s=sta[c^1][k];
     48                 ll num=dp[c^1][k];
     49                 int p=(s>>bit[j-1])&3,q=(s>>bit[j])&3;
     50                 if (!mp[i][j])
     51                 {
     52                     if (!p&&!q) add(s,num);
     53                 }
     54                 else if (!p&&!q)
     55                 {
     56                     if (!mp[i+1][j]||!mp[i][j+1]) continue;
     57                     s+=(1<<bit[j-1])+(1<<(bit[j]+1)) ;
     58                     add(s,num);
     59                 }
     60                 else if (!p&&q){
     61                     if (mp[i][j+1]) add(s,num);
     62                     if (mp[i+1][j]){
     63                         s+=(1<<bit[j-1])*q-(1<<bit[j])*q;
     64                         add(s,num);
     65                     }
     66                 }
     67                 else if (!q&&p){
     68                     if (mp[i+1][j]) add(s,num);
     69                     if (mp[i][j+1]){
     70                         s+=(1<<bit[j])*p-(1<<bit[j-1])*p;
     71                         add(s,num);
     72                     }
     73                 }
     74                 else if (p+q==2){
     75                     int b=1;
     76                     for (int t=j+1;t<=m;t++){
     77                         int v=(s>>bit[t])&3;
     78                         if (v==1) b++;
     79                         if (v==2) b--;
     80                         if (!b){
     81                             s-=(1<<bit[t]);
     82                             break;
     83                         }
     84                     }
     85                     s-=(1<<bit[j-1])+(1<<bit[j]);
     86                     add(s,num);
     87                 }
     88                 else if (p+q==4){
     89                     int b=1;
     90                     for (int t=j-2;t>=0;t--){
     91                         int v=(s>>bit[t])&3;
     92                         if (v==2) b++;
     93                         if (v==1) b--;
     94                         if (!b){
     95                             s+=(1<<bit[t]);
     96                             break;
     97                         }
     98                     }
     99                     s-=2*(1<<bit[j-1])+2*(1<<bit[j]);
    100                     add(s,num);
    101                 }
    102                 else if (p==2&&q==1){
    103                     s-=(1<<(bit[j-1]+1))+(1<<bit[j]);
    104                     add(s,num);
    105                 }
    106                 else if (p==1&&q==2){
    107                     if (i==ex&&j==ey) ans+=num;
    108                 }
    109             }
    110         }
    111     }
    112 }
    113 int main(){
    114     for (int i=0;i<=25;i++)
    115         bit[i]=i<<1;
    116     scanf("%d%d",&n,&m);
    117     tot[0]=1; dp[c][1]=1;
    118     for (int i=1;i<=n;i++){
    119         scanf("%s",s+1);
    120         for (int j=1;j<=m;j++)
    121             if (s[j]=='.') mp[i][j]=1,ex=i,ey=j;
    122     }
    123     DP();
    124     printf("%lld
    ",ans);
    125 }

    那么对于这道题目,分析其特殊性质。

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  • 原文地址:https://www.cnblogs.com/fengzhiyuan/p/8486624.html
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