题目链接：http://acm.fzu.edu.cn/problem.php?pid=1977
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Problem Description The pollution of the earth is so serious that people can not survive any more. Fortunately, people have found a new planet that maybe has life, and we call it "Pandora Planet". Leonardo Da Vinci is the only astronaut on the earth. He will be sent to the Pandora Planet to gather some plant specimens and go back. The plant specimen is important to the people to decide whether the planet is fit to live or not. Assuming that Da Vinci can only move in an N×M grid. The positions of the plant specimens he wants to collect are all marked by the satellite. His task is to find a path to collect all the plant specimens and return to the spaceship. There are some savage beasts in the planet. Da Vinci can not investigate the grid with the savage beast. These grids are also marked by the satellite. In order to save time Da Vinci could only visit each grid exactly once and also return to the start grid, that is, you can not visit a grid twice except the start grid. You should note that you can choose any grid as the start grid. Now he wants to know the number of different paths he can collect all the plant specimens. We only care about the path and ignore where the start grid is, so the two paths in Figure 1 are considered as the same. Figure 1 Input The first line of the input contains an integer T (T≤100), indicating the number of cases. Each case begins with a line containing two integers N and M (1≤N, M≤12), the size of the planet is N×M. Each of the following N lines contains M characters Gij(1≤i≤N, 1≤j≤M), Gij denotes the status of the grid in row i and column j, where 'X' denotes the grid with savage beast, '*' denotes the safe grid that you can decide to go or not, 'O' denotes the plant specimen you should collect. We guarantee that there are at least three plant specimens in the map. Output For each test case, print a line containing the test case number (beginning with 1) and the number of different paths he can collect all the plant specimens. You can make sure that the answer will fit in a 64-bit signed integer. Sample Input 2 2 2 OO O* 4 4 ***O XO** **O* XX** Sample Output Case 1: 1 Case 2: 7
题意是这样的，给你一个地图，其中有些地方必须经过，有些地方不必须经过，有些地方不能经过，求满足的回路数有多少条。在逐格递推的时候统计已经过plant specimens的个数，在和并连通分量的时候，如果plant specimens的个数已经达到总个数，在判断此时状态的环的个数，如果只有一个，那么加入ans，并且不计入下一个格子的状态。
1 //STATUS:C++_AC_359MS_19832KB 2 #include<stdio.h> 3 #include<stdlib.h> 4 #include<string.h> 5 #include<math.h> 6 #include<iostream> 7 #include<string> 8 #include<algorithm> 9 #include<vector> 10 #include<queue> 11 #include<stack> 12 #include<map> 13 using namespace std; 14 #define LL long long 15 #define pii pair<int,int> 16 #define Max(a,b) ((a)>(b)?(a):(b)) 17 #define Min(a,b) ((a)<(b)?(a):(b)) 18 #define mem(a,b) memset(a,b,sizeof(a)) 19 #define lson l,mid,rt<<1 20 #define rson mid+1,r,rt<<1|1 21 const int N=15,INF=0x3f3f3f3f,MOD=4001,STA=8000010; 22 const double DNF=1e13; 23 24 int g[N][N],code[N],ma[N]; 25 int T,n,m,ex,ey,sumo,cnto; 26 LL ans; 27 28 struct Hash{ //Hash表,MOD为表长,STA为表大小 29 int first[MOD],next[STA],size; 30 LL f[STA],sta[STA]; 31 void init(){ 32 size=0; 33 mem(first,-1); 34 } 35 void add(LL st,LL ans){ 36 int i,u=st%MOD; 37 for(i=first[u];i!=-1;i=next[i]){ 38 if(sta[i]==st){ 39 f[i]+=ans; 40 return; 41 } 42 } 43 sta[size]=st; 44 f[size]=ans; 45 next[size]=first[u]; 46 first[u]=size++; 47 } 48 }hs[2]; 49 50 void shift(int p) //换行移位 51 { 52 int k; 53 LL sta; 54 for(k=0;k<hs[!p].size;k++){ 55 sta=hs[!p].sta[k]<<3; 56 hs[p].add(sta,hs[!p].f[k]); 57 } 58 } 59 60 LL getsta() //最小表示法 61 { 62 LL i,cnt=1,sta=0; 63 mem(ma,-1); 64 ma[0]=0; 65 for(i=0;i<=m;i++){ 66 if(ma[code[i]]==-1)ma[code[i]]=cnt++; 67 code[i]=ma[code[i]]; 68 sta|=(LL)code[i]<<(3*i); 69 } 70 return sta; 71 } 72 73 void getcode(LL sta) 74 { 75 int i; 76 for(i=0;i<=m;i++){ 77 code[i]=sta&7; 78 sta>>=3; 79 } 80 } 81 82 void unblock(int i,int j,int p) 83 { 84 int k,t; 85 LL cnt,x,y; 86 for(k=0;k<hs[!p].size;k++){ 87 getcode(hs[!p].sta[k]); 88 x=code[j],y=code[j+1]; 89 cnt=hs[!p].f[k]; 90 if(x && y){ //合并连通分量 91 code[j]=code[j+1]=0; 92 if(x!=y){ 93 for(t=0;t<=m;t++) 94 if(code[t]==y)code[t]=x; 95 hs[p].add(getsta(),cnt); 96 } 97 else if(cnto>=sumo){ 98 int ok=1; 99 for(t=0;t<=m;t++){ 100 if(code[t]){ok=0;break;} 101 } 102 if(ok)ans+=cnt; 103 } 104 } 105 else if(x&&!y || !x&&y){ //延续连通分量 106 t=x?x:y; 107 if(g[i+1][j]){ 108 code[j]=t;code[j+1]=0; 109 hs[p].add(getsta(),cnt); 110 } 111 if(g[i][j+1]){ 112 code[j]=0;code[j+1]=t; 113 hs[p].add(getsta(),cnt); 114 } 115 } 116 else { //创建新连通分量 117 if(g[i][j]==1) 118 hs[p].add(getsta(),cnt); 119 if(g[i+1][j] && g[i][j+1]){ 120 code[j]=code[j+1]=8; 121 hs[p].add(getsta(),cnt); 122 } 123 } 124 } 125 } 126 127 void block(LL j,int p) 128 { 129 int k; 130 for(k=0;k<hs[!p].size;k++){ 131 getcode(hs[!p].sta[k]); 132 code[j]=code[j+1]=0; 133 hs[p].add(getsta(),hs[!p].f[k]); 134 } 135 } 136 137 LL slove() 138 { 139 int i,j,p; 140 ans=0;cnto=0; 141 hs[0].init(); 142 hs[p=1].init(); 143 hs[0].add(0,1); 144 for(i=0;i<n;i++){ 145 for(j=0;j<m;j++){ 146 if(g[i][j]){ 147 cnto+=g[i][j]==2; 148 unblock(i,j,p); 149 } 150 else block(j,p); 151 hs[p=!p].init(); 152 } 153 shift(p); //换行移位 154 hs[p=!p].init(); 155 } 156 return ans; 157 } 158 159 int main() 160 { 161 // freopen("in.txt","r",stdin); 162 int i,j,k=1; 163 char c; 164 scanf("%d",&T); 165 while(T--) 166 { 167 scanf("%d%d",&n,&m); 168 sumo=0; 169 mem(g,0); 170 for(i=0;i<n;i++){ 171 for(j=0;j<m;j++){ 172 scanf(" %c",&c); 173 if(c=='O'){ 174 g[i][j]=2; 175 sumo++; 176 } 177 else if(c=='*')g[i][j]=1; 178 else g[i][j]=0; 179 } 180 } 181 182 slove(); 183 184 printf("Case %d: %I64d\n",k++,ans); 185 } 186 return 0; 187 }