洛谷1002 容斥原理+dfs OR DP

Posted 2020-09-24 SiriusRen

//By SiriusRen
#include <bits/stdc++.h>
using namespace std;
#define int long long
int n,m,sx,sy,xx[]={1,1,2,2,-1,-1,-2,-2,0},yy[]={2,-2,1,-1,2,-2,1,-1,0};
int stk1[25],stk2[25],stk[25],C[45][45],Ans,top;
bool check(int x,int y){return x>=0&&y>=0&&x<=n&&y<=m;}
int solve(int dep){
    int lastx=0,lasty=0,ans=1;
    for(int i=1;i<=dep;i++){
        int tx=stk1[stk[i]]-lastx,ty=stk2[stk[i]]-lasty;
//        printf("tx=%d ty=%d\n",tx,ty);
        ans*=C[tx+ty][tx];
        lastx=stk1[stk[i]],lasty=stk2[stk[i]];
    }
    int tx=n-lastx,ty=m-lasty;
    return ans*C[tx+ty][tx];
}
void dfs(int x,int y,int nw,int dep){
    if(dep&1)Ans-=solve(dep);
    else Ans+=solve(dep);
//    printf("x=%lld y=%lld dep=%lld Ans=%lld\n",x,y,dep,solve(dep));
    for(int i=1;i<=top;i++){
        if(i!=nw&&stk1[i]>=x&&stk2[i]>=y){
            stk[dep+1]=i;
            dfs(stk1[i],stk2[i],i,dep+1);
        }
    }
}
signed main(){
    scanf("%lld%lld%lld%lld",&n,&m,&sx,&sy);
    for(int i=0;i<=40;i++){
        C[i][0]=C[i][i]=1;
        for(int j=1;j<i;j++)
            C[i][j]=C[i-1][j-1]+C[i-1][j];
    }
    for(int i=0;i<=8;i++){
        int dx=sx+xx[i],dy=sy+yy[i];
        if(check(dx,dy))stk1[++top]=dx,stk2[top]=dy;
    }
    dfs(0,0,0,0);
    printf("%lld\n",Ans);
}

//By SiriusRen
#include <stdio.h>
long long n,m,sx,sy,i,j,vis[25][25],F[25][25],xx[]={1,1,2,2,-1,-1,-2,-2,0},yy[]={2,-2,1,-1,2,-2,1,-1,0};
int check(int x,int y){return x>=0&&y>=0&&x<=n&&y<=m;}
signed main(){
    scanf("%I64d%I64d%I64d%I64d",&n,&m,&sx,&sy);
    for(i=0;i<=8;i++){
        int dx=sx+xx[i],dy=sy+yy[i];
        if(check(dx,dy))vis[dx][dy]=1;
    }
    F[0][0]=1;
    for(i=0;i<=n;i++){
        for(j=0;j<=m;j++)if(!vis[i][j]){
            if(i)F[i][j]+=F[i-1][j];
            if(j)F[i][j]+=F[i][j-1];
        }
    }printf("%I64d\n",F[n][m]);
    return 0;
}