codeforces CF37E Trial for Chief BFS最短路
E. Trial for Chief
Having unraveled the Berland Dictionary,
the scientists managed to read the notes of the chroniclers of that time.
For example, they learned how the chief of the ancient Berland tribe was chosen.
As soon as enough pretenders was picked, the following test took place among them:
the chief of the tribe took a slab divided by horizontal and vertical stripes into identical squares
(the slab consisted of $ N $ lines and $ M $ columns)
and painted every square black or white.
Then every pretender was given a slab of the same size but painted entirely white.
Within a day a pretender could paint any side-linked set of the squares of the slab some color.
The set is called linked if for any two squares belonging to the set
there is a path belonging the set on which any two neighboring squares share a side.
The aim of each pretender is to paint his slab in the exactly the same way as the chief’s slab is painted.
The one who paints a slab like that first becomes the new chief.
Scientists found the slab painted by the ancient Berland tribe chief.
Help them to determine the minimal amount of days
needed to find a new chief if he had to paint his slab in the given way.
Input
The first line contains two integers $ N $ and $ M (1 ≤ N, M ≤ 50) $ — the number of lines and columns on the slab.
The next $ N $ lines contain $ M $ symbols each — the final coloration of the slab.
$ W $ stands for the square that should be painted white and $ B $ — for the square that should be painted black.
Output
In the single line output the minimal number of repaintings of side-linked areas
needed to get the required coloration of the slab.
Examples
input1
3 3
WBW
BWB
WBW
output1
2
input2
2 3
BBB
BWB
output2
1
题目大意
-
要把一块大小为 $ N \times M $ 的白木板的某些格子涂成黑色。
-
每次可以选择一个连通块，全涂白或者全涂黑，问最少要涂几次。
-
$ N, M \le 50 $
题解
-
每个点与周围颜色相同的格子连边权为 $ 0 $ 的边
-
每个点与周围颜色不同的格子连边权为 $ 1 $ 的边
-
求每个点到最远的黑格子的最短路，取 $ min+1 $ 就是答案。
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define inf 1e9+7
#define maxn 2510
int fx[]={0,0,1,-1};
int fy[]={1,-1,0,0};
struct edge{ int v,w,nxt; }e[maxn<<2];
int n,m,head[maxn],tot,dis[maxn],col[60][60],pot[60][60],bk[maxn],ans=inf;
char ch[maxn];
bool vis[maxn],fb;
void add(int u,int v,int w){ e[++tot].v=v; e[tot].w=w; e[tot].nxt=head[u]; head[u]=tot; }
void bfs(int s){
memset(dis,0x3f,sizeof(dis)); memset(vis,0,sizeof(vis)); queue<int>q;
q.push(s); dis[s]=0; vis[s]=1;
while(!q.empty()){
int u=q.front(); q.pop(); vis[u]=0;
for(int i=head[u];i;i=e[i].nxt)
if(dis[e[i].v]>dis[u]+e[i].w){
dis[e[i].v]=dis[u]+e[i].w;
if(!vis[e[i].v]){ vis[e[i].v]=1; q.push(e[i].v); }
}
}
}
int main(){
scanf("%d %d",&n,&m);
for(int i=1;i<=n;++i){
scanf("%s",ch);
for(int j=1;j<=m;++j){
col[i][j]=(ch[j-1]=='B');
fb|=col[i][j];
pot[i][j]=++pot[0][0];
if(col[i][j]) bk[++bk[0]]=pot[i][j];
}
}
if(!fb){ puts("0"); return 0; }
for(int i=1;i<=n;++i)
for(int j=1;j<=m;++j)
for(int nx,ny,k=0;k<4;++k){
nx=i+fx[k]; ny=j+fy[k];
if(nx<1||nx>n||ny<1||ny>m) continue;
add(pot[i][j],pot[nx][ny],col[i][j]!=col[nx][ny]);
}
for(int i=1;i<=n;++i)
for(int j=1;j<=m;++j){
bfs(pot[i][j]);
int temp=0;
for(int j=1;j<=bk[0];++j) temp=max(temp,dis[bk[j]]);
ans=min(ans,temp);
}
printf("%d",ans+1);
return 0;
}
/*
# 40098602
When 2018-07-09 05:31:27
Who PotremZ
Problem E - Trial for Chief
Lang GNU C++
Verdict Accepted
Time 436 ms
Memory 200 KB
*/
