A destructive fire raged through the Forest, and several trees were damaged by it. Precisely speaking, you have a n×mn×m rectangle map which represents the damaged part of the Forest. The damaged trees were marked as "X" while the remaining ones were marked as ".". You are sure that all burnt trees are shown on the map. All the trees outside the map are undamaged.
The firemen quickly extinguished the fire, and now they are investigating the cause of it. The main version is that there was an arson: at some moment of time (let's consider it as 00) some trees were set on fire. At the beginning of minute 00, only the trees that were set on fire initially were burning. At the end of each minute, the fire spread from every burning tree to each of 88 neighboring trees. At the beginning of minute TT, the fire was extinguished.
The firemen want to find the arsonists as quickly as possible. The problem is, they know neither the value of TT (how long the fire has been raging) nor the coordinates of the trees that were initially set on fire. They want you to find the maximum value of TT (to know how far could the arsonists escape) and a possible set of trees that could be initially set on fire.
Note that you'd like to maximize value TT but the set of trees can be arbitrary.
The first line contains two integer nn and mm (1≤n,m≤1061≤n,m≤106, 1≤n⋅m≤1061≤n⋅m≤106) — the sizes of the map.
Next nn lines contain the map. The ii-th line corresponds to the ii-th row of the map and contains mm-character string. The jj-th character of the ii-th string is "X" if the corresponding tree is burnt and "." otherwise.
It's guaranteed that the map contains at least one "X".
In the first line print the single integer TT — the maximum time the Forest was on fire. In the next nn lines print the certificate: the map (n×mn×m rectangle) where the trees that were set on fire are marked as "X" and all other trees are marked as ".".
3 6 XXXXXX XXXXXX XXXXXX
1 ...... .X.XX. ......
10 10 .XXXXXX... .XXXXXX... .XXXXXX... .XXXXXX... .XXXXXXXX. ...XXXXXX. ...XXXXXX. ...XXXXXX. ...XXXXXX. ..........
2 .......... .......... ...XX..... .......... .......... .......... .....XX... .......... .......... ..........
4 5 X.... ..XXX ..XXX ..XXX
0 X.... ..XXX ..XXX ..XXX
题意：给一个n*m<=1e6的图，"X"代表着火，"."代表没着火，火只在这个范围里面蔓延。着火的格子每回合会蔓延到周围的8个格子，给出了最后的火情图，求着火最多多少回合，并构造刚着火的图。
想法：很明显蔓延后的火都是一个矩形，所以我先横着遍历1遍，再竖着遍历遍，基本可以确认这个火最多蔓延了多少回合，构造图通过前缀和来判断，比较简单。然后我就被这组数据卡住了
7 5 ..XXX ..XXX ..XXX .XXX. XXX.. XXX.. XXX..
我一开始的输出
1 ..... ...X. ..... ..... ..... .X... .....
答案
0 ..XXX ..XXX ..XXX .XXX. XXX.. XXX.. XXX..
被这个数据卡住是因为，横着扫一遍竖着扫一遍只能确定最多蔓延回合不会超过这个数值，但是构造出来的图不一定能蔓延回原来的图。
做法：二分答案，先记录终图的前缀和，然后构造起始图，该点为"X"意味着终图中以该点为中心的m范围全为"X"，用二维前缀和来判断。否则为"."。构造的同时记录起始图的前缀和。
构造完起始图后我们再检查初始是否合法，同样是用前缀和，该点为"X"意味着起始图中以该点为中心的m范围有"X"或者该点为"."意味着起始图中以该点为中心的m范围。
分析复杂度，每次二分我们需要构造1次和检查1次，复杂度是2e6。对于1e6的图需要二分20次复杂度是4e7，分析可得，蔓延最大不会超过1000回合，只需二分10次，复杂度变2e7。而横着扫一遍竖着扫一遍可以确定二分上界，这个操作复杂度是2e6，这样比纯二分快点。。
#include <bits/stdc++.h> using namespace std; const int N=2e6+5; char p[N]; char v[N]; int u[N]; int z[N]; int n,m; inline bool solve(int zd) { for(int i=1;i<=n;++i) { for(int j=1;j<=m;++j) { int t=i*(m+1)+j; v[t]='.'; if(p[t]=='X') { int flag=0; int rx=i+zd; int ry=j+zd; int lx=i-zd; int ly=j-zd; if(rx>n||ry>m||lx<=0||ly<=0)flag=1; if((u[rx*(m+1)+ry]-u[rx*(m+1)+ly-1]-u[(lx-1)*(m+1)+ry])+u[(lx-1)*(m+1)+ly-1]!=((zd*2+1)*(zd*2+1)))flag=1; if(!flag)v[t]='X'; } z[t]=(v[t]=='X'?1:0)+z[t-1]+z[t-m-1]-z[t-m-2]; } } for(int i=1;i<=n;++i) { for(int j=1;j<=m;++j) { int t=i*(m+1)+j; if(p[t]=='X') { int rx=i+zd; int ry=j+zd; int lx=i-zd; int ly=j-zd; if(rx>n||ry>m||lx<=0||ly<=0)continue; if((z[rx*(m+1)+ry]-z[rx*(m+1)+ly-1]-z[(lx-1)*(m+1)+ry]+z[(lx-1)*(m+1)+ly-1])==0)return false; } } } return true; } int main() { scanf("%d%d",&n,&m); for(int i=1;i<=n;++i) { for(int j=1;j<=m;++j) { int t=i*(m+1)+j; scanf(" %c",&p[t]); u[t]=(p[t]=='X'?1:0)+u[t-1]+u[t-m-1]-u[t-m-2]; } } int zd=10000005; for(int i=1;i<=n;++i) { for(int j=1;j<=m;++j) { int t=0; while(p[i*(m+1)+j]=='X') { j++; t++; if(j==m+1)break; } if(t&&zd>t)zd=t; } } for(int i=1;i<=m;++i) { for(int j=1;j<=n;++j) { int t=0; while(p[i+j*(m+1)]=='X') { j++; t++; if(j==n+1)break; } if(t&&zd>t)zd=t; } } zd=(zd-1)/2; if(zd==0) { printf("0 "); for(int i=1;i<=n;++i) { for(int j=1;j<=m;++j)printf("%c",p[i*(m+1)+j]); printf(" "); } return 0; } int l=0,r=zd; int ans=0; while(l<=r) { int m=(l+r)>>1; if(solve(m)) { l=m+1; ans=m; } else r=m-1; } printf("%d ",ans); solve(ans); for(int i=1;i<=n;++i) { for(int j=1;j<=m;++j)printf("%c",v[i*(m+1)+j]); printf(" "); } return 0; }
总结：直接开二维数组不可取，我是用了一维的来表示二维的，用vector应该会方便不少。然后就是要多练，此外对题目的复述要准确，问人时完全描述成了另一道题。。这都是值得努力的地方