c/c++语言开发共享cf232E. Quick Tortoise(分治 bitset dp)

题意

题目链接

sol

感觉这个思路还是不错的

#include<bits/stdc++.h> using namespace std; const int maxn = 501, ss = 5e6 + 10; inline int read() {     char c = getchar(); int x = 0, f = 1;     while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();     return x * f; }  int n, m, q, ans[ss]; char s[maxn][maxn]; bitset<maxn> f[maxn][maxn], g[maxn][maxn], empty; struct query {     int x1, y1, x2, y2, id; }q[ss]; void solve(int l, int r, vector<query> q) {     if(l > r) return ;     vector<query> lq, rq;     int mid = l + r >> 1;     //f[i][j]从i,j能到达的mid中的点集     //g[i][j]从mid能到达i, j的点集      for(int i = mid; i >= l; i--) {         for(int j = m; j >= 1; j--) {             f[i][j] = empty;             if(i == mid) f[i][j][j] = (s[i][j] == '.');             if(i + 1 <= mid && s[i + 1][j] == '.') f[i][j] = f[i][j] | f[i + 1][j];             if(j + 1 <= m && s[i][j + 1] == '.') f[i][j] = f[i][j] | f[i][j + 1];                      }     }     for(int i = mid; i <= r; i++) {         for(int j = 1; j <= m; j++) {             g[i][j] = empty;             if(i == mid) g[i][j][j] = (s[i][j] == '.');             if(i - 1 >= mid && s[i - 1][j] == '.') g[i][j] = g[i][j] | g[i - 1][j];             if(j - 1 >= 1 && s[i][j - 1] == '.') g[i][j] = g[i][j] | g[i][j - 1];         }     }     for(auto &cur : q) {         if(cur.x2 < mid) lq.push_back(cur);         else if(cur.x1 > mid) rq.push_back(cur);         else {             //cout << f[cur.x1][cur.y1] << endl;             //cout << g[cur.x2][cur.y2] << endl;             ans[cur.id] = (f[cur.x1][cur.y1] & g[cur.x2][cur.y2]).count();         }     }     solve(l, mid - 1, lq);     solve(mid + 1, r, rq); } int main() {     n = read(); m = read();     for(int i = 1; i <= n; i++) scanf("%s", s[i] + 1);     q = read();     vector<query> now;     for(int i = 1; i <= q; i++) {         q[i].x1 = read(), q[i].y1 = read(), q[i].x2 = read(), q[i].y2 = read(); q[i].id = i;         now.push_back(q[i]);     }     solve(1, n, now);     for(int i = 1; i <= q; i++) puts(ans[i] ? "yes" : "no");     return 0; } /* 3 3 ... .## .#. 1 1 1 3 1   3 3 ... .## .#. 5 1 1 3 3 1 1 1 3 1 1 3 1 1 1 1 2 1 1 2 1 */