打模板的经验：
1.变量名取一样，换行也一样，不要宏定义
2.大小写，少写，大括号
#include<algorithm>
#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<stdio.h>
#include<vector>
#include<queue>
#include<string>
#include<ctime>
#include<stack>
#include<map>
#include<set>
#include<list>
using namespace std;
#define rep(i,j,k) for(int i = (int)j;i <= (int)k;i ++)
#define per(i,j,k) for(int i = (int)j;i >= (int)k;i --)
#define debug(x) cerr<<#x<<" = "<<(x)<<endl
#define mmm(a,b) memset(a,b,sizeof(a))
#define pb push_back
//#define x first
//#define y second
typedef double db;
typedef long long ll;
const int MAXD = 15;
const int STATE = 1e6 + 5;
const int HASH = 3e5 + 7;
const int MOD = 1e9 + 7;
int N, M, K;
int maze[MAXD][MAXD];
int code[MAXD];
int ch[MAXD];
int num;
struct HASHMAP
{
int head[HASH], next[STATE], size;
ll state[STATE];
int f[STATE];
void init()
{
size = 0;
mmm(head, -1);
}
void push(ll 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;
}
}
ll encode(int *code, int m)
{
int cnt = 1;
mmm(ch, -1);
ch[0] = 0;
ll st = 0;
rep(i, 0, m)
{
if (ch[code[i]] == -1)ch[code[i]] = cnt++;
code[i] = ch[code[i]];
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;
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 dplock(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()
{
cin >> N >> M >> K;
mmm(maze, 0);
rep(i, 1, N)
{
scanf("%s", &str);
for (int j = 1; j <= M; j++)
if (str[j - 1] == '.')
maze[i][j] = 1;
}
}
void solve()
{
int cur = 0;
hm[cur].init();
hm[cur].push(0, 1);
rep(i, 1, N)
rep(j, 1, M)
{
hm[cur ^ 1].init();
if (maze[i][j])dpblank(i, j, cur);
else dplock(i, j, cur);
cur ^= 1;
}
int ans = 0;
for (int i = 0; i < hm[cur].size; i++)
if (hm[cur].state[i] == K)
{
ans += hm[cur].f[i];
ans %= MOD;
}
printf("%d
", ans);
}
int main()
{
int t;
cin >> t;
while (t--)
{
init();
solve();
}
//cin >> t;
}
/*
2
4 4 1
**..
....
....
....
4 1
....
....
....
....
*/
/* 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[code[i]]==-1)ch[code[i]]=cnt++; code[i]=ch[code[i]]; 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 ",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 4 4 1 **.. .... .... .... 4 1 .... .... .... .... Sample Output 6 */
#include<algorithm>#include<iostream>#include<stdlib.h>#include<string.h>#include<math.h>#include<stdio.h>#include<vector>#include<queue>#include<string>#include<ctime>#include<stack>#include<map>#include<set>#include<list>using namespace std;#define rep(i,j,k) for(int i = (int)j;i <= (int)k;i ++)#define per(i,j,k) for(int i = (int)j;i >= (int)k;i --)#define debug(x) cerr<<#x<<" = "<<(x)<<endl#define mmm(a,b) memset(a,b,sizeof(a))#define pb push_back//#define x first//#define y second
typedef double db;typedef long long ll;
const int MAXD = 15;const int STATE = 1e6 + 5;const int HASH = 3e5 + 7;const int MOD = 1e9 + 7;int N, M, K;int maze[MAXD][MAXD];int code[MAXD];int ch[MAXD];int num;struct HASHMAP{int head[HASH], next[STATE], size;ll state[STATE];int f[STATE];void init(){size = 0;mmm(head, -1);}void push(ll 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;}}ll encode(int *code, int m){int cnt = 1;mmm(ch, -1);ch[0] = 0;ll st = 0;rep(i, 0, m){if (ch[code[i]] == -1)ch[code[i]] = cnt++;code[i] = ch[code[i]];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;
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 dplock(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(){cin >> N >> M >> K;mmm(maze, 0);rep(i, 1, N){scanf("%s", &str);for (int j = 1; j <= M; j++)if (str[j - 1] == '.')maze[i][j] = 1;}}void solve(){int cur = 0;hm[cur].init();hm[cur].push(0, 1);rep(i, 1, N)rep(j, 1, M){hm[cur ^ 1].init();if (maze[i][j])dpblank(i, j, cur);else dplock(i, j, cur);cur ^= 1;}int ans = 0;for (int i = 0; i < hm[cur].size; i++)if (hm[cur].state[i] == K){ans += hm[cur].f[i];ans %= MOD;}printf("%d
", ans);
}int main(){int t;cin >> t;while (t--){init();solve();}//cin >> t;}/*24 4 1**..............4 1................*/