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「CF1006F」Xor-Paths Meet-in-the-middle

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字数统计: 551字 | 阅读时长 ≈ 3分钟

(第一份屯题计划?)

Links there:CF1006F

题意

给定一个$n \times m$的矩阵,从起点$(1,1)$起步到$(n,m)$,每次只能向下或者向右走.

求所有满足路径上的异或和为 $k$ 的路径条数.

$ n,m \leq 20 ,\space k\leq 1e18$

思路

采用中间相遇的方法.显然每次从$(1,1)$ 到 $(n,m)$ 需要走$(n+m-2)$步.前一半暴力大法师.后一半直接判断是否可以异或到$k$即可.$k$很大的话不如开个$map$.

Code

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//my vegetable has exploded. :(
#include<bits/stdc++.h>
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MM(x,y) memset(x,y,sizeof(x))
#define MCPY(a,b) memcpy(a,b,sizeof(b))
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,b,a) for(int i=b;i>=a;i--)
#define fi first
#define se second
#define int long long
using namespace std;

inline int quickpow(int m,int n,int p){int b=1;while(n){if(n&1)b=b*m%p;n=n>>1;m=m*m%p;}return b;}
inline int getinv(int x,int p){return quickpow(x,p-2,p);}
inline int read(void){
	int x=0,f=1;char ch=getchar();
	while(!isdigit(ch)){f=ch=='-'?-1:1;ch=getchar();}
	while(isdigit(ch)){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x * f;
}
const int MAXN = 25;
map<int,int>cnt[MAXN][MAXN];
int a[MAXN][MAXN],ans=0,n,m,k,h;
inline void dfs(int x,int y,int val,int step){
    if (x < 1 || x > n || y < 1 || y > m) return ;
    int cur = val ^ a[x][y];
    if (x+y == h+2) {
         ++cnt[x][y][cur];
        return ;
    }
    dfs(x+1,y,cur,step+1);
    dfs(x,y+1,cur,step+1);
}
inline void dfs2(int x,int y,int val,int step){
    if (x < 1 || x > n || y < 1 || y > m) return ;
    int cur = val ^ a[x][y];
    if (x+y == h+2) {
        if (cnt[x][y][val^k]) ans += cnt[x][y][val^k];
        return ;
    }
    dfs2(x-1,y,cur,step+1);
    dfs2(x,y-1,cur,step+1);
}
///------------------head------------------
signed main(signed argc, char *argv[]){
    n=read(),m=read(),k=read();
    h=(n+m-2)>>1;
    rep(i,1,n) rep(j,1,m) a[i][j]=read();
    dfs(1,1,0,0);
    dfs2(n,m,0,0);
    printf("%lld\n",ans);
	return 0;
}
/* Examples: */
/*
3 3 11
2 1 5
7 10 0
12 6 4
*/

/*
3
*/