洛谷1002 容斥原理+dfs OR DP

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

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