https://icpcarchive.ecs.baylor.edu/external/79/7958.pdf
7958 Maximum Islands
You are mapping a faraway planet using
a satellite.
Your satellite has captured an image of the planet’s surface. The photographed section can be modeled as a
grid. Each grid cell is either land, water,
or covered by clouds. Clouds mean that
the surface could either be land or water, but we can’t tell.
An island is a set of connected land cells. Two cells are considered connected if they share an edge.
Given the image, determine the maximum number of islands that is consistent with the given
information.
Input
The input fle contains several test cases, each of them as described below.
The frst line contains two space-separated integers nand m (1≤ n; m ≤40).
Each of the next n lines containsm characters, describing the satellite image. Land cells are denoted
by ‘L’, water cells are denoted by ‘W’, and cells covered by clouds are denoted by ‘C’.
Output
For each test case, print, on a single line, a single integer indicating the maximum number of islands
that is consistent with the given grid.
Sample Input
5 4
LLWL
CCCC
CCCC
CCCC
LWLL
Sample Output
8
给你一个地图，L表示陆地，W表示水，C表示你可以任意安排，问最多有多少个陆地联通块。
先把所有已知陆地联通块dfs。接着，把不和陆地相邻的所有C区域安排为陆地，最优情况肯定是一个格子为一片陆地。
要构造二分图，就要黑白染色。染色之后，把相邻的C互相连边，对所有C组成的图求最大独立集。
当然，不黑白染色也可以，只是这时会有重边，数组要开够。
说道这里我就要吐槽了！这题亲测，数据范围应该是N<=60,然而题目里却是40，好坑啊！
导致我调到现在，WA了二十几次。
#include <cstdio>
#include <iostream>
#include <string.h>
#include <string>
#include <map>
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <math.h>
#include <cmath>
#include <stack>
#define mem0(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
using namespace std;
typedef long long ll;
typedef long double ld;
const int maxn=65,inf=0x3f3f3f3f;
const ll llinf=0x3f3f3f3f3f3f3f3f;
const ld pi=acos(-1.0L);
int match[maxn*maxn];
int p[maxn][maxn],head[maxn*maxn];
bool visit[maxn][maxn],v[maxn*maxn];
char s[maxn][maxn];
int dir[4][2];
int n,m,num;
struct Edge {
int from,to,pre;
};
Edge edge[maxn*maxn*2];
void addedge(int from,int to) {
edge[num]=(Edge){from,to,head[from]};
head[from]=num++;
edge[num]=(Edge){to,from,head[to]};
head[to]=num++;
}
void dfs(int i,int j) {
visit[i][j]=1;
for (int k=0;k<4;k++) {
int x=i+dir[k][0],y=j+dir[k][1];
if (x>0&&y>0&&x<=n&&y<=m)
if (!visit[x][y]&&s[x][y]=='L') dfs(x,y);
if (s[x][y]=='C') s[x][y]='W';
}
}
bool hungary(int now) {
for (int i=head[now];i!=-1;i=edge[i].pre) {
int to=edge[i].to;
if (!v[to]) {
v[to]=1;
if (!match[to]||hungary(match[to])) {
match[to]=now;match[now]=to;
return true;
}
}
}
return false;
}
int main() {
dir[0][0]=dir[1][0]=dir[2][1]=dir[3][1]=0;
dir[0][1]=dir[2][0]=1;dir[1][1]=dir[3][0]=-1;
int i,j,k,ans=0;
scanf("%d%d",&n,&m);
for (i=1;i<=n;i++) {
scanf("%s",s[i]+1);
}
mem0(visit);
for (i=1;i<=n;i++)
for (j=1;j<=m;j++)
if (!visit[i][j]&&s[i][j]=='L') dfs(i,j),ans++;
int cnt=0;
memset(head,-1,sizeof(head));
num=0;
for (i=1;i<=n;i++)
for (j=1;j<=m;j++)
if (s[i][j]=='C')
p[i][j]=++cnt;
for (i=1;i<=n;i++)
for (j=1;j<=m;j++)
if (s[i][j]=='C'&&(i+j)%2==0)
for (k=0;k<4;k++) {
int x=i+dir[k][0],y=j+dir[k][1];
if (x>0&&y>0&&x<=n&&y<=m)
if (s[x][y]=='C') addedge(p[i][j],p[x][y]);
}
int sum=0;
mem0(match);
for (i=1;i<=cnt;i++)
if (!match[i]) {
mem0(v);v[i]=1;
if (hungary(i)) sum++;
}
ans+=cnt-sum;
printf("%d\n",ans);
return 0;
}