70分TLE！（求调）

andy22 @ 2024-07-19 20:21:19

#include <bits/stdc++.h>
using namespace std;

const int N = 1e3+5;
char mp[N][N];
bool visit[N][N];
int n, m;
int dx[] = {-1, 0, 1,  0};
int dy[] = {0,  1, 0, -1};
bool check(int x, int y){
    return (x >= 0 && x < n && y >= 0 && y < n);
}
int bfs(int x, int y) {
    int ans = 1;
    visit[x][y] = 1;
    queue<pair<int,int>> que;
    que.push({x, y});
    while(!que.empty()){
        x = que.front().first;
        y = que.front().second;
        que.pop();
//      cout << ans;
        for(int i = 0; i < 4; i++) {
            int nx = x + dx[i];
            int ny = y + dy[i];
            if(check(nx,ny) && !visit[nx][ny] && mp[x][y] != mp[nx][ny]) {
                ans++;
//              cout << nx << " " << ny << endl;
                visit[nx][ny] = 1;
                que.push({nx,ny});
            }
        }
    }
    return ans;
}
int main() {
    scanf("%d %d", &n, &m);
    for(int i = 0; i < n; i++) {
        scanf("%s", &mp[i]);
    }
    int x, y;
    for(int i = 1; i <= m; i++) {
        scanf("%d %d", &x, &y);
        memset(visit, 0, sizeof(visit));
        printf("%d\n", bfs(x, y));
    }

    return 0;
}

by han1219 @ 2024-07-19 20:52:30

先开long long试试

by ATION001 @ 2024-07-19 20:55:29

@andy22 温馨提示：本题用普通思想会超时，要预处理。

代码

#include<bits/stdc++.h>
using namespace std;
using pii=pair<int,int>;
int _x[4]={0,0,-1,1},_y[4]={-1,1,0,0};
int b[1005][1005],sum[100010];
char a[1005][1005];
int n,m;
void bfs(int x,int y,int flag){
    queue<pii>q;
    q.push({x,y});
    sum[flag]++,b[x][y]=flag;
    while(q.size()){
        pii p=q.front();
        q.pop();
        for(int i=0;i<4;i++){
            int dx=p.first+_x[i],dy=p.second+_y[i];
            if(a[dx][dy]=='o'||b[dx][dy]!=-1||a[p.first][p.second]==a[dx][dy]){
                continue;
            }
            sum[flag]++,b[dx][dy]=flag;
            q.push({dx,dy});
        }
    }
}
int main(){
    fill(a[0],a[1005],'o');
    fill(b[0],b[1005],-1);
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            cin>>a[i][j];
        }
    }
    for(int i=1,x,y;i<=m;i++){
        cin>>x>>y;
        if(b[x][y]==-1){
            bfs(x,y,i);
        }else{
            sum[i]=sum[b[x][y]];
        }
    }
    for(int i=1;i<=m;i++){
        cout<<sum[i]<<endl; 
    }
    return 0;
}

求关。

by ATION001 @ 2024-07-19 21:33:15

把地图当做连通块预处理。