Codeforces 1064 D

D - Labyrinth

对于位置(i,j), j - c = R - L = const(常数), 其中R表示往右走了几步,L表示往左走了几步

所以R越大, L就越大, R越小, L就越小, 所以只需要最小化L和R中的其中一个就可以了

由于每次变化为0或1,所以用双端队列写bfs, 保证最前面的值最小, 简化版的dijkstra

不过看到好多没写双端队列的也过了......

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 2e3 + 5;
char s[N][N];
int mnr[N][N];
int dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};
deque<piii> q;
int n, m;
void bfs(int x, int y, int rx, int ry) {
    mem(mnr, 0x3f);
    mnr[x][y] = 0;
    q.push_back({{x, y}, 0});
    while(!q.empty()) {
        piii p = q.front();
        q.pop_front();
        for (int i = 0; i < 4; i++) {
            int xx = p.fi.fi + dir[i][0];
            int yy = p.fi.se + dir[i][1];
            if(i == 1) {
                if(1 <= xx && xx <= n && 1 <= yy && yy <= m && s[xx][yy] == '.' && p.se + 1 < mnr[xx][yy]) {
                    mnr[xx][yy] = p.se+1;
                    q.push_back({{xx, yy}, p.se+1});
                }
            }
            else {
                if(1 <= xx && xx <= n && 1 <= yy && yy <= m && s[xx][yy] == '.' && p.se < mnr[xx][yy]) {
                    mnr[xx][yy] = p.se;
                    q.push_front({{xx, yy}, p.se});
                }
            }
        }
    }
}
int main() {
    int r, c, x, y;
    scanf("%d %d", &n, &m);
    scanf("%d %d", &r, &c);
    scanf("%d %d", &x, &y);
    for (int i = 1; i <= n; i++) scanf("%s", s[i]+1);
    int ans = 0;
    bfs(r, c, x, y);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            int cst = j - c;
            int l = mnr[i][j] - cst;
            if(mnr[i][j] <= y && l <= x) ans++;
        }
    }
    printf("%d
", ans);
    return 0;
}