并查集+bfs+暴力滑窗 Codeforces Round #356 (Div. 2) E

http://codeforces.com/contest/680/problem/E

题目大意：给你一个n*n的图，然后图上的 . （我们下面都叫做‘点’）表示可以走，X表示不能走，你有如下的操作，每次你可以选择一个k*k的框，把其中的所有的X都变成 ‘点’，问在该操作后点相连的数目最多是多少？

没看别人代码和思路自己思考并优化了好久，于是敲了两个多小时。。。还是代码能力太差了

思路：当然最最单纯的做法就是O(n^4)。为了优化，起初想用二维树状数组+并查集的，结果发现窗口中的‘点‘’容易造成重复计算，于是换了种方法，每次事先统计‘点’和X，然后再每次暴力枚举边就好了。复杂度O(n*n*4*4*k*log)

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\n")
const int maxn =  + ;
int par[maxn * maxn], cnt[maxn * maxn];
bool vis[maxn * maxn];
int n, k;
char ch[maxn][maxn];
int dx[] = {, , -, };
int dy[] = {, , , -};

int bfs(int x, int y, int fa){
    queue<pair<int, int> > que;
    que.push(mk(x, y));
    int cnt = ;
    while (!que.empty()){
        pair<int, int> u = que.front(); que.pop();
        for (int i = ; i < ; i++){
            int nx = u.fi + dx[i], ny = u.se + dy[i];
            if (nx <=  || ny <=  || nx > n || ny > n) continue;
            if (ch[nx][ny] != '.' || vis[nx * n + ny]) continue;
            vis[nx * n + ny] = true;
            cnt++;
            par[nx * n + ny] = fa;
            que.push(mk(nx, ny));
        }
    }
    return cnt;
}

int pointcnt[maxn * maxn];
inline int init(int i){
    for (int x = ; x <= n; x++)
        for (int y = ; y <= n; y++)
            pointcnt[x * n + y] = vis[x * n + y] = ;

    int cntx = ;
    for (int x = i; x <= k + i - ; x++)///take care of border
        for (int y = ; y < k; y++){///最后一列先不放进去
            if (ch[x][y] == '.') pointcnt[par[x * n + y]]++;
            if (ch[x][y] == 'X') cntx++;
        }
    return cntx;
}

int solve(){
    int ans = ;
    for (int i = ; i <= n - k + ; i++){
        int cntx = init(i);
        for (int y = ; y <= n - k + ; y++){///四个方向的遍历
            ///clear上一列的,insert目前列的
            for (int x = i; x <= i + k - ; x++){
                if (ch[x][y - ] == '.')  pointcnt[par[x * n + y - ]]--;
                else if (ch[x][y - ] == 'X') cntx--;
                if (ch[x][y + k - ] == '.') pointcnt[par[x * n + y + k - ]]++;
                else if (ch[x][y + k - ] == 'X') cntx++;
            }
            int pin = ;
            vector<int> v;
            ///左右列, 变行
            for (int x = i; x <= i + k - ; x++){
                for (int j = ; j < ; j++){
                    int nx = x + dx[j], ny = y + dy[j];
                    if (nx <=  || ny <=  || nx > n || ny > n) continue;
                    if (ch[nx][ny] != '.') continue;
                    int pos = par[nx * n + ny];
                    if (vis[pos]) continue;
                    vis[pos] = true; v.push_back(pos); pin += pointcnt[pos];
                }
                for (int j = ; j < ; j++){
                    int nx = x + dx[j], ny = y + k -  + dy[j];
                    if (nx <=  || ny <=  || nx > n || ny > n) continue;
                    if (ch[nx][ny] != '.') continue;
                    int pos = par[nx * n + ny];
                    if (vis[pos]) continue;
                    vis[pos] = true; v.push_back(pos); pin += pointcnt[pos];
                }
            }
            ///上下行，变列
            for (int yy = y; yy <= y + k - ; yy++){
                for (int j = ; j < ; j++){
                    int nx = i + dx[j], ny = yy + dy[j];
                    if (nx <=  || ny <=  || nx > n || ny > n) continue;
                    if (ch[nx][ny] != '.') continue;
                    int pos = par[nx * n + ny];
                    if (vis[pos]) continue;
                    vis[pos] = true; v.push_back(pos); pin += pointcnt[pos];
                }
                for (int j = ; j < ; j++){
                    int nx = i + k -  + dx[j], ny = yy + dy[j];
                    if (nx <=  || ny <=  || nx > n || ny > n) continue;
                    if (ch[nx][ny] != '.') continue;
                    int pos = par[nx * n + ny];
                    if (vis[pos]) continue;
                    vis[pos] = true; v.push_back(pos); pin += pointcnt[pos];
                }
            }
            int len = v.size();
            int tmp = k * k - pin;
            for (int i = ; i < len; i++){
                tmp += cnt[v[i]];
                vis[v[i]] = false;
            }
            ans = max(ans, tmp);
        }
    }
    return ans;
}

int main(){
    scanf("%d%d", &n, &k);
    for (int i = ; i <= n; i++) scanf("%s", ch[i] + );
    for (int i = ; i <= n; i++)
        for (int j = ; j <= n; j++)
            par[i * n + j] = i * n + j;
    for (int i = ; i <= n; i++){
        for (int j = ; j <= n; j++){
            if (ch[i][j] == '.' && !vis[i * n + j]){
                vis[i * n + j] = true;
                cnt[i * n + j] = bfs(i, j, i * n + j);
            }
        }
    }
    printf("%d\n", solve());
    return ;
}

然后敲完之后跑出来是2600ms，然后有人是300ms左右就过了的，具体思路差不多，大概就是我初始化+vector多了一堆没用的东西导致复杂度变得很高了。

不过我也看了一下他们的代码发现，果然差距还是好大

代码来源：http://blog.csdn.net/Fsss_7/article/details/51706546#reply

#include<bits/stdc++.h>
using namespace std;
const int N=;
typedef long long ll;
typedef unsigned long long ull;
char s[N];
int n,a[*N][*N],b[N][N];
int g=,q[N*N],f[N*N],xm[N*N],ym[N*N],xx[N*N],yx[N*N];

///这里应该就是单纯的跑一下，然后知道目前联通块的边界
void dfs(int x,int y) {
    b[x][y]=g;f[g]++;///标记是第几个联通快，和记录着联通块的数目
    xm[g]=min(xm[g],x);xx[g]=max(xx[g],x);
    ym[g]=min(ym[g],y);yx[g]=max(yx[g],y);
    if (x>&&b[x-][y]==-) dfs(x-,y);
    if (x<n&&b[x+][y]==-) dfs(x+,y);
    if (y>&&b[x][y-]==-) dfs(x,y-);
    if (y<n&&b[x][y+]==-) dfs(x,y+);

}

int main(){
    int i,j,h,k,tot,ans=;
    scanf("%d%d", &n, &k);
    for (i=;i<=n;i++) {
        scanf("%s", s);
        for (j=;j<=n;j++)
        if (s[j-]=='.') b[i][j]=-;///存在点
        else {
            b[i][j]=;///不是点
            a[i][j]+=;a[i][j+k]+=-;///然后标记障碍物所对应的方框
            a[i+k][j]+=-;a[i+k][j+k]+=;
        }
    }
    memset(xm,0x7f,sizeof(xm));
    memset(ym,0x7f,sizeof(ym));
    for (i=;i<=n;i++)
        for (j=;j<=n;j++)
            if (b[i][j]==-) {
                g++,dfs(i,j);
                if (xx[g]-xm[g]+<=k&&yx[g]-ym[g]+<=k) {///边界超过K的就不会被k*k包围啦，然后对边界进行处理
                    a[xx[g]][yx[g]]+=f[g];  a[xm[g]+k][ym[g]+k]+=f[g];//左上右下
                    a[xx[g]][ym[g]+k]-=f[g];  a[xm[g]+k][yx[g]]-=f[g];//左下右上
                }
            }

    for (int i = ; i <= n; i++){
        for (int j = ; j <= n; j++){
            printf("%d ", a[i][j]);
        }
        cout << endl;
    }

    for (i=;i<=n;i++)///a[i][j]是统计(i,j)~(i-k, j-k)这个矩形中，只在内部的‘点’和X的数目
        for (j=;j<=n;j++)
            a[i][j]+= a[i][j-] + a[i-][j] - a[i-][j-];
    for (i=k;i<=n;i++)
        for (j=k;j<=n;j++) {///枚举(k,k)开始的端点
            tot=;
            for (h=j-k+;h<=j;h++)///左列
                if (!q[b[i-k][h]]) q[b[i-k][h]]=,tot+=f[b[i-k][h]];///用q标记该连通块是否被访问过，然后b所记录的是第几个块的标号
            for (h=j-k+;h<=j;h++)///右列
                if (!q[b[i+][h]]) q[b[i+][h]]=,tot+=f[b[i+][h]];
            for (h=i-k+;h<=i;h++)///上行
                if (!q[b[h][j-k]]) q[b[h][j-k]]=,tot+=f[b[h][j-k]];
            for (h=i-k+;h<=i;h++)///下行
                if (!q[b[h][j+]]) q[b[h][j+]]=,tot+=f[b[h][j+]];
            ans=max(ans,tot+a[i][j]);
            printf("tot = %d a[%d][%d] = %d\n", tot, i, j, a[i][j]);
            ///printf("a[%d][%d] = %d\n", i, j, a[i][j]);
            ///初始化
            for (h=j-k+;h<=j;h++) q[b[i-k][h]]=;
            for (h=j-k+;h<=j;h++) q[b[i+][h]]=;
            for (h=i-k+;h<=i;h++) q[b[h][j-k]]=;
            for (h=i-k+;h<=i;h++) q[b[h][j+]]=;
        }
    printf("%d\n", ans);
    return ;
}