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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#72239#4788. GravitylmeowdnTL 2ms5512kbC++201.8kb2023-01-15 10:44:512023-01-15 10:44:54

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-01-15 10:44:54]
  • 评测
  • 测评结果:TL
  • 用时:2ms
  • 内存:5512kb
  • [2023-01-15 10:44:51]
  • 提交

answer

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define fi first
#define se second
#define eb emplace_back
#define bp __builtin_parity
#define y1 yyl
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<pii> vp;
typedef bitset<1009> bset;

int read() {
	int x=0,w=1; char c=getchar();
	while(!isdigit(c)) {if(c=='-') w=-1; c=getchar();}
	while(isdigit(c)) {x=x*10+c-'0'; c=getchar();}
	return x*w;
}

const int N=2002,inf=0x3f3f3f3f;
int n,m,id[N][N],tot;
vector<vector<pair<int,short>>> e;
vector<short> f;
char s[N][N];
set<pair<short,int> >q;

void dfs(int x,int y,int p) {
	if(!x||!y||x>n||y>m||s[x][y]!='#'||id[x][y]) return;
	id[x][y]=p;
	dfs(x+1,y,p), dfs(x,y+1,p), dfs(x-1,y,p), dfs(x,y-1,p);
}

signed main() {
	n=read(), m=read();
	rep(i,1,n) scanf("%s",s[i]+1);
	rep(i,1,n) rep(j,1,m) if(s[i][j]=='#'&&!id[i][j]) dfs(i,j,++tot);
	++tot;
	rep(j,1,m) id[n+1][j]=tot;
	f.resize(tot+1); e.resize(tot+1);
	rep(j,1,m) {
		int lst=n+1;
		per(i,n,1) if(s[i][j]=='#') {
			if(id[lst][j]!=id[i][j]) {
				e[id[lst][j]].eb(make_pair(id[i][j],lst-i-1));
			}
			lst=i;
		}
	}
	rep(i,1,tot-1) f[i]=n+1;
	q.insert(make_pair(0,tot));
	while(q.size()) {
		int u=(*q.begin()).se; q.erase(q.begin());
		for(auto ed:e[u]) {
			int v=ed.fi, w=ed.se;
			if(f[v]>f[u]+w) {
				if(f[v]==n+1) q.insert(make_pair(-f[u]-w,v));
				else q.erase(make_pair(-f[v],v)), q.insert(make_pair(-f[u]-w,v));
				f[v]=f[u]+w;
			}
		}
	}
	per(i,n,1) rep(j,1,m) {
		if(s[i][j]=='#') s[i][j]='.', s[i+f[id[i][j]]][j]='#';
	}
	rep(i,1,n) {
		rep(j,1,m) putchar(s[i][j]);
		puts("");
	}
	return 0;
}

Details

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Test #1:

score: 100
Accepted
time: 2ms
memory: 5512kb

input:

10 10
..........
..######..
..#....#..
..#.#..#..
..#..#.#..
..#....#..
..######..
..........
..#....#..
.......#..


output:

..........
..........
..######..
..#....#..
..#....#..
..#....#..
..#.##.#..
..######..
.......#..
..#....#..

result:

ok 10 lines

Test #2:

score: -100
Time Limit Exceeded

input:

1583 1799
#..###..##..#.####.##.##.###..#.....##..#.#.#.#......#.....##..#.##...#.#....#..##.###...#...##.###.#.....#.##.###...#..##.#...###..#.###.#...###..#.......#...#....#.#..#...##........#.#..#..##.....###...#..#.####..####...#..##......#........#..#.##.##..#..#....##.##.##..#..##.....###....#...

output:


result: