CodeForces-1006F Path 双向DFS

传送门：CodeForces-1006F

思路分析

一次DFS的复杂度是$2^40$，两次的话是$2*2^20$
从头和从尾开始，跑到相同点的时候统计答案

样例输入

3 3 11
2 1 5
7 10 0
12 6 4

样例输出

3

AC代码

#include <bits/stdc++.h>
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;

int n,m;
ll a[22][22];
map<ll,ll>dp[22][22];

void dfs1(int x,int y,ll s){
    if(x>n||x<1||y>m||y<1) return ;
    s^=a[x][y];
    if(x+y==(n+m)/2+1) {
        dp[x][y][s]++;
        return ;
    }
    dfs1(x+1,y,s);
    dfs1(x,y+1,s);
}

ll dfs2(int x,int y,ll s){
    if(x>n||x<1||y>m||y<1) return 0;
    if(x+y==(n+m)/2+1) return dp[x][y][s];
    s^=a[x][y];
    return dfs2(x-1,y,s)+dfs2(x,y-1,s);
}

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    ll k;
    read(n,m,k);
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) read(a[i][j]);
    dfs1(1,1,0);
    cout<<dfs2(n,m,k)<<endl;



    return 0;
}