cf232E. Quick Tortoise(分治 bitset dp)

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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
*/

cf232E. Quick Tortoise(分治 bitset dp)

标签：else   技术   技术分享   https   ack   mes   tor   def   %s   

原文地址：https://www.cnblogs.com/zwfymqz/p/10273942.html