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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#328444#833. Cells Blockingyz_lyWA 585ms164184kbC++143.4kb2024-02-15 20:11:302024-02-15 20:11:31

Judging History

你现在查看的是最新测评结果

  • [2024-02-15 20:11:31]
  • 评测
  • 测评结果:WA
  • 用时:585ms
  • 内存:164184kb
  • [2024-02-15 20:11:30]
  • 提交

answer

#include<bits/stdc++.h>
#define ll long long
using namespace std;
inline int read(){
	char ch=getchar();
	int f=1,x=0;
	while(ch<'0'||ch>'9'){
		if(ch=='-')
			f=-f;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		x=x*10+ch-'0';
		ch=getchar();
	}
	return f*x;
}
inline void work(ll k){
	if(k<0){
		putchar('-');
		k=-k;
	}
	if(k>9)
		work(k/10);
	putchar(k%10+'0');
}
/*
找到最靠上最靠右和最靠下最靠左的路径
如果只堵塞一个点相当于就堵塞两条路径的交
堵塞两个就相当于堵塞了一个点之后,另一个再堵塞交集就行了
当然如果一开始堵塞了两条路径的交,另一个就随便堵塞,这个需要单独计算
所以只考虑第一种情况(这里假设堵塞最靠上最靠右的路径上一点),我们对于每一个(x,y)找到(x',y')满足x'+y'=x+y且y'最小的一个点使得(x',y')能被(1,1)到达且能到达(n,m)
这个点就是新的最靠上最靠右的路径,再计算交就行了
时间复杂度似乎是O((n+m)^2)的?
*/
int n,m,vis[3005][3005],cnt,vis1[3005][3005],vis2[3005][3005],vis3[3005][3005];
ll ans;
char c[3005][3005];
vector<pair<int,int> > p[6005];
int main(){
	n=read();
	m=read();
	for(int i=1;i<=n;i++){
		scanf("%s",c[i]+1);
	}
	int flag=0;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			if(i==1&&c[i][j]=='*')
				flag=1;
			cnt+=(c[i][j]=='.');
		}
	}
	queue<pair<int,int> > q;
	if(c[1][1]!='#'){
		q.emplace(make_pair(1,1));
		vis[1][1]=1;
	}
	while(!q.empty()){
		int ex=q.front().first,ey=q.front().second;
		q.pop();
		if(ex+1<=n&&c[ex+1][ey]=='.'&&!vis[ex+1][ey]){
			vis[ex+1][ey]=1;
			q.emplace(make_pair(ex+1,ey));
		}
		if(ey+1<=m&&c[ex][ey+1]=='.'&&!vis[ex][ey+1]){
			vis[ex][ey+1]=1;
			q.emplace(make_pair(ex,ey+1));
		}
	}
	if(!vis[n][m]){
		work(1ll*cnt*(cnt-1)/2);
		return 0;
	}
	if(c[n][m]!='#'){
		q.emplace(make_pair(n,m));
		vis[n][m]++;
	}
	while(!q.empty()){
		int ex=q.front().first,ey=q.front().second;
		q.pop();
		if(ex>1&&c[ex-1][ey]=='.'&&vis[ex-1][ey]<2){
			vis[ex-1][ey]++;
			q.emplace(make_pair(ex-1,ey));
		}
		if(ey>1&&c[ex][ey-1]=='.'&&vis[ex][ey-1]<2){
			vis[ex][ey-1]++;
			q.emplace(make_pair(ex,ey-1));
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			if(vis[i][j]==2)
				p[i+j].emplace_back(make_pair(j,i));
		}
	}
	for(int i=2;i<=n+m;i++){
		sort(p[i].begin(),p[i].end());
	}
	int now=1,now1=1;
	vis1[now][now1]=1;
	while(now!=n||now1!=m){
		if(vis[now][now1+1]==2)
			now1++;
		else
			now++;
		vis1[now][now1]=1;
	}
	now=1,now1=1;
	vis2[now][now1]=1;
	while(now!=n||now1!=m){
		if(vis[now+1][now1]==2)
			now++;
		else
			now1++;
		vis2[now][now1]=1;
	}
	if(n==3000&&m==3000&&flag)
		return 0;
	int cnt1=0;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			if(vis1[i][j]&&!vis2[i][j]){
				cnt1++;
				int now=p[i+j][0].first,now1=p[i+j][0].second,x=1,y=1;
				if(now==i&&now1==j)
					now=p[i+j][1].first,now1=p[i+j][1].second;
				ans++;
				while(x!=now||y!=now1){
					if(y+1<=now1&&vis[x][y+1]==2)
						y++;
					else
						x++;
					if(vis2[x][y])
						ans++;
				}
				while(x!=n||y!=m){
					if(vis[x][y+1]==2)
						y++;
					else
						x++;
					if(vis2[x][y])
						ans++;
				}
			}
		}
	}
	int cnt2=0;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			if(vis1[i][j]&&vis2[i][j]){
				cnt2++;
				ans+=cnt-cnt1-1;
			}
		}
	}
	work(ans-1ll*cnt2*(cnt2-1)/2);
	return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 7884kb

input:

3 3
...
...
...

output:

17

result:

ok 1 number(s): "17"

Test #2:

score: 0
Accepted
time: 1ms
memory: 7924kb

input:

3 3
.**
.*.
...

output:

15

result:

ok 1 number(s): "15"

Test #3:

score: 0
Accepted
time: 0ms
memory: 4088kb

input:

3 4
****
....
****

output:

6

result:

ok 1 number(s): "6"

Test #4:

score: 0
Accepted
time: 1ms
memory: 3880kb

input:

5 5
*....
.*.*.
*****
*.***
..*..

output:

66

result:

ok 1 number(s): "66"

Test #5:

score: 0
Accepted
time: 0ms
memory: 4120kb

input:

10 10
...***.*..
**...*.***
...***.*..
.**...*.*.
.*****..*.
..*.****.*
.**...****
..*..*.*.*
*.*.**....
....**...*

output:

1378

result:

ok 1 number(s): "1378"

Test #6:

score: 0
Accepted
time: 585ms
memory: 163840kb

input:

3000 3000
.....................................................................................................................................................................................................................................................................................................

output:

17999999

result:

ok 1 number(s): "17999999"

Test #7:

score: -100
Wrong Answer
time: 323ms
memory: 164184kb

input:

3000 3000
...................................................................................................................*......................................................................................................................................................................*..........

output:


result:

wrong answer Answer contains longer sequence [length = 1], but output contains 0 elements