插头DP,多条回路求回路数。
可以当作模版来记。注重理解轮廓线。
#include <iostream> #include <string.h> #include <iostream> #include <stdio.h> using namespace std; const int HASH=10007; const int STATE=5000; //状态数 const int MAXD=15; int code[MAXD],maze[MAXD][MAXD]; int N,M; struct HASHMAP { int head[HASH],next[STATE],state[STATE],size; long long f[STATE]; void init() { size=0; memset(head,-1,sizeof(head)); } void push(int st,long long ans) { int i,h=st%HASH; for(i=head[h];i!=-1;i=next[i]) if(st==state[i]) { f[i]+=ans; return; } f[size]=ans; state[size]=st; next[size]=head[h]; head[h]=size++; } }hm[2]; void decode(int *code,int m,int st) { int i; for(i=m;i>=0;i--) { code[i]=st&1; st>>=1; } } int encode(int *code,int m) { int i,st=0; for(i=0;i<=m;i++) { st<<=1; st|=code[i]; } return st; } void init() { scanf("%d%d",&N,&M); for(int i=1;i<=N;i++) for(int j=1;j<=M;j++) scanf("%d",&maze[i][j]); for(int i=1;i<=N;i++)maze[i][M+1]=0; for(int i=1;i<=M;i++)maze[N+1][i]=0;//边界补0 } void shift(int *code,int m)//换行 { int i; for(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)//左边和上边都有插头 { code[j-1]=code[j]=0; if(j==M)shift(code,M); hm[cur^1].push(encode(code,M),hm[cur].f[k]); } else if(left||up) { if(maze[i][j+1]) { code[j-1]=0; code[j]=1; hm[cur^1].push(encode(code,M),hm[cur].f[k]); } if(maze[i+1][j]) { code[j-1]=1; code[j]=0; if(j==M)shift(code,M); hm[cur^1].push(encode(code,M),hm[cur].f[k]); } } else { if(maze[i+1][j]&&maze[i][j+1]) { code[j]=code[j-1]=1; hm[cur^1].push(encode(code,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; if(j==M)shift(code,M); hm[cur^1].push(encode(code,M),hm[cur].f[k]); } } void solve() { int i,j,cur; long long ans=0; 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; } for(i=0;i<hm[cur].size;i++) ans+=hm[cur].f[i]; printf("There are %I64d ways to eat the trees.\n",ans); } int main() { /*freopen("in.txt","r",stdin); freopen("out.txt","w",stdout);*/ int T; int iCase=0; scanf("%d",&T); while(T--) { iCase++; printf("Case %d: ",iCase); init(); solve(); } return 0; }
HDU 1693 Eat the Trees,布布扣,bubuko.com
原文地址:http://blog.csdn.net/imutzcy/article/details/25708019