"Fzu" 1977 Pandora Adventure

http://acm.fzu.edu.cn/problem.php?pid=1977

Test instructions: Nxm grid, there are 3 kinds of lattice, ' O ' must Pass, ' * ' can choose to pass, ' X ' cannot pass. Now requires that the path passes through all ' O ' and is the number of simple loops

#include <cstdio> #include <cstring> #include <algorithm>using namespace Std;typedef long ll; struct H {static const int m=799997;struct E {int next, to;} E[m];int Ihead, CNT, Hash[m];ll sum[m]; H () {ihead=cnt=0; memset (hash,-1, sizeof hash); memset (sum, 0, sizeof sum);} BOOL Find (int x, int &pos) {pos=x%m;while (1) {if (hash[pos]==x) return False, if (hash[pos]==-1) break; ++pos; if (pos== M) pos=0; }hash[pos]=x; return true;} void ins (int a, ll b) {int pos; if (!find (A, POS)) {sum[pos]+=b; return;} E[++cnt].next=ihead; ihead=cnt; E[cnt].to=pos; Sum[pos]=b;} void CLR () {for (int i=ihead; i; i=e[i].next) hash[e[i].to]=-1, sum[e[i].to]=0; ihead=cnt=0;}} T1, T2; #define BIT (b) ((a) << ((b) <<1)) #define The CLR (A, a) (A^= ((a) &bit (3,b))) #define GET (A, B) (3& ( A) >> ((b) <<1))) const int N=15;int N, M, All;ll ans;int mp[n][n];int Find (int s, int col, int flag) {int Sum=0;i F (flag) {for (int i=col; i<=m; ++i) {int K=get (s, i); if (k==1) ++sum;if (k==2)--sum;if (!sum) return i;} else {for (int i=col; i>=0;-I.) {int K=get (s, i), if (k==1)--sum;if (k==2) ++sum;if (!sum) return i;}} return-1;} void print (int s) {for (int i=0; i<=m; ++i) {int K=get (s, i), if (k==0) Putchar (' # '), else if (k==1) Putchar (' ('); else I  F (k==2) Putchar (') '); }puts ("");} #define F puts ("error"), bool Next (int s, int row, int col, bool U, BOOL D, BOOL L, bool R, int &t) {if (row==n-1&& AMP;D) | | (row==0&&u) | | (Col==m-1&&r) | | (col==0&&l)) Return 0;if ((D&&!mp[row+1][col]) | | (R&&!mp[row][col+1])) Return 0;int L=get (S, col), U=get (S, col+1), Flag=get (S, m+1), D=0, R=0;if ((l&&! L) | | (!l&&l) | | (u&&! U) | | (!u&&u)) return 0;//printf ("state:"); print (s);//printf ("row:%d, col:%d, u:%d, d:%d, l:%d, r:%d", Row, col, U, D, L, R);//printf ("Left:"); if (l==1) printf ("("); if (l==2) printf (")"); if (l==0) printf ("#");//printf ("UPTP:"); if (u==1) printf ("("); if (u==2) printf (")"); if (u==0) printf ("#"); Puts (""); t=s; CLR (T, col); CLR (t, col+1), if (!l &&!u) {if (R && D) d=1, r=2;} else if (l && u) {if (l==1 && u==1) {int Pos=find (s, col+1, 1); CLR (T, POS); T|=bit (1, POS);} else if (l==2 && u==2) {int Pos=find (s, col, 0); CLR (T, POS); T|=bit (2, POS);} else if (l==1 && u==2) flag=1;} else if (l &&!u) {if (D) d=l, R=0;if (R) d=0, R=l;} else if (!l && u) {if (D) D=u, R=0;if (R) d=0, R=u;} T|=bit (d, Col); T|=bit (R, col+1); if (col==m-1) t<<=2;  Puts ("=============\nnext"); Print (t); t&=all;if (flag) t|=bit (1, m+1); return 1;} void BFs () {H *q1, *q2;q1=&t1; q2=&t2;q1->clr (); q2->clr (); q1->ins (0, 1); for (int row=0; row<n; ++row ) for (int col=0; col<m; ++col) {q2->clr ();//printf ("q1->cnt:%d\n", q1->cnt); for (int i=q1->ihead; i; i= Q1->e[i].next) {int s=q1->hash[q1->e[i].to], T, Flag=get (S, m+1);//print (s); ll Sum=q1->sum[q1->e[i]. To];if (mp[row][col]==2 && flag) continue;if (mp[row][col]<=1) {if (next (S, Row, col, 0, 0, 0, 0, T)) q2->ins (t, sum);} if (mp[row][col]>=1 &&!flag) {//Forgot to award flag again for a long time ... if (next (S, Row, col, 1, 1, 0, 0, T)) q2->ins (t, sum); if (next (s, Row, col, 1, 0, 1, 0, T)) Q2->ins (t, sum), if (next (S, Row, col, 1, 0, 0, 1, t)) q2->ins (t, sum), if (next (S, Row, col, 0, 1, 1, 0, T)) Q2->ins (t, sum), if (next (S, Row, col, 0, 1, 0, 1, t)) q2->ins (t, sum), if (next (S, Row, col, 0, 0, 1, 1, T)) q2->ins (t, sum);} }swap (Q1, Q2);} for (int i=q1->ihead; i; i=q1->e[i].next) ans+=q1->sum[q1->e[i].to];} int main () {int cs; scanf ("%d", &cs), char s[n];for (int cc=1; cc<=cs; ++cc) {ans=0;scanf ("%d%d", &n, &m); al L=bit (1, m+1) -1;for (int i=0; i<n; ++i) {scanf ("%s", s), for (int j=0; j<m; ++j) {if (s[j]== ' * ') mp[i][j]=1;else if (S[j] = = ' O ') Mp[i][j]=2;else mp[i][j]=0;}} BFS ();p rintf ("Case%d:%i64d\n", CC, ans);} return 0;}

　　

Today, the ability to commit to Max?

There is no need to consider the situation is not necessary to expand the status Qaq

The topic is actually a template problem. It's just that we don't know where the ring is linked, so let's open another State to indicate whether the current contour is already a ring.

"Fzu" 1977 Pandora Adventure