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 ") const int maxn = 500 + 5; int par[maxn * maxn], cnt[maxn * maxn]; bool vis[maxn * maxn]; int n, k; char ch[maxn][maxn]; int dx[] = {1, 0, -1, 0}; int dy[] = {0, 1, 0, -1}; int bfs(int x, int y, int fa){ queue<pair<int, int> > que; que.push(mk(x, y)); int cnt = 1; while (!que.empty()){ pair<int, int> u = que.front(); que.pop(); for (int i = 0; i < 4; i++){ int nx = u.fi + dx[i], ny = u.se + dy[i]; if (nx <= 0 || ny <= 0 || 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 = 1; x <= n; x++) for (int y = 1; y <= n; y++) pointcnt[x * n + y] = vis[x * n + y] = 0; int cntx = 0; for (int x = i; x <= k + i - 1; x++)///take care of border for (int y = 1; 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 = 0; for (int i = 1; i <= n - k + 1; i++){ int cntx = init(i); for (int y = 1; y <= n - k + 1; 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 - 1; x++){ if (ch[x][y - 1] == '.') pointcnt[par[x * n + y - 1]]--; else if (ch[x][y - 1] == 'X') cntx--; if (ch[x][y + k - 1] == '.') pointcnt[par[x * n + y + k - 1]]++; else if (ch[x][y + k - 1] == 'X') cntx++; } int pin = 0; vector<int> v; ///左右列, 变行 for (int x = i; x <= i + k - 1; x++){ for (int j = 0; j < 4; j++){ int nx = x + dx[j], ny = y + dy[j]; if (nx <= 0 || ny <= 0 || 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 = 0; j < 4; j++){ int nx = x + dx[j], ny = y + k - 1 + dy[j]; if (nx <= 0 || ny <= 0 || 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 - 1; yy++){ for (int j = 0; j < 4; j++){ int nx = i + dx[j], ny = yy + dy[j]; if (nx <= 0 || ny <= 0 || 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 = 0; j < 4; j++){ int nx = i + k - 1 + dx[j], ny = yy + dy[j]; if (nx <= 0 || ny <= 0 || 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 = 0; 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 = 1; i <= n; i++) scanf("%s", ch[i] + 1); for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) par[i * n + j] = i * n + j; for (int i = 1; i <= n; i++){ for (int j = 1; 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 ", solve()); return 0; }
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=510; typedef long long ll; typedef unsigned long long ull; char s[N]; int n,a[2*N][2*N],b[N][N]; int g=0,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>1&&b[x-1][y]==-1) dfs(x-1,y); if (x<n&&b[x+1][y]==-1) dfs(x+1,y); if (y>1&&b[x][y-1]==-1) dfs(x,y-1); if (y<n&&b[x][y+1]==-1) dfs(x,y+1); } int main(){ int i,j,h,k,tot,ans=0; scanf("%d%d", &n, &k); for (i=1;i<=n;i++) { scanf("%s", s); for (j=1;j<=n;j++) if (s[j-1]=='.') b[i][j]=-1;///存在点 else { b[i][j]=0;///不是点 a[i][j]+=1;a[i][j+k]+=-1;///然后标记障碍物所对应的方框 a[i+k][j]+=-1;a[i+k][j+k]+=1; } } memset(xm,0x7f,sizeof(xm)); memset(ym,0x7f,sizeof(ym)); for (i=1;i<=n;i++) for (j=1;j<=n;j++) if (b[i][j]==-1) { g++,dfs(i,j); if (xx[g]-xm[g]+1<=k&&yx[g]-ym[g]+1<=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 = 1; i <= n; i++){ for (int j = 1; j <= n; j++){ printf("%d ", a[i][j]); } cout << endl; } for (i=1;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=1;j<=n;j++) a[i][j]+= a[i][j-1] + a[i-1][j] - a[i-1][j-1]; for (i=k;i<=n;i++) for (j=k;j<=n;j++) {///枚举(k,k)开始的端点 tot=0; for (h=j-k+1;h<=j;h++)///左列 if (!q[b[i-k][h]]) q[b[i-k][h]]=1,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+1;h<=j;h++)///右列 if (!q[b[i+1][h]]) q[b[i+1][h]]=1,tot+=f[b[i+1][h]]; for (h=i-k+1;h<=i;h++)///上行 if (!q[b[h][j-k]]) q[b[h][j-k]]=1,tot+=f[b[h][j-k]]; for (h=i-k+1;h<=i;h++)///下行 if (!q[b[h][j+1]]) q[b[h][j+1]]=1,tot+=f[b[h][j+1]]; ans=max(ans,tot+a[i][j]); printf("tot = %d a[%d][%d] = %d ", tot, i, j, a[i][j]); ///printf("a[%d][%d] = %d ", i, j, a[i][j]); ///初始化 for (h=j-k+1;h<=j;h++) q[b[i-k][h]]=0; for (h=j-k+1;h<=j;h++) q[b[i+1][h]]=0; for (h=i-k+1;h<=i;h++) q[b[h][j-k]]=0; for (h=i-k+1;h<=i;h++) q[b[h][j+1]]=0; } printf("%d ", ans); return 0; }