题解 启程的日子
一个我以为的转化是将同色连通块缩点
那么就是求最少用多少个白点可以将所有黑点连通
然而一个格子可以被加两次再减一次，所以就假了
那么观察（？）发现操作次数 \(\leqslant 3\)，证明见 3 时的构造方法
为 0 和 1 直接算黑连通块个数即可
为 2 检查是否存在一个白连通块连通了所有黑点
为 3 需要一些构造：
前两次操作以这样一个划分为基础
第一次操作将黄，绿 +1
第二次将红，蓝 +1
第三次将除棕色部分外全部 -1
正确性显然，鬼知道怎么想到的
点击查看代码
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 510
#define fir first
#define sec second
#define ll long long
//#define int long long
int n, m;
char s[N][N];
bool vis[N*N], rev;
map<int, bool> to[N*N];
set<pair<int, int>> pts[N*N];
int id[N][N], dsu[N*N], tot, cnt;
const int dlt[][2]={{-1,0},{1,0},{0,1},{0,-1}};
inline int find(int p) {return dsu[p]==p?p:dsu[p]=find(dsu[p]);}
inline void uni(int s, int t) {
s=find(s), t=find(t);
if (s==t) return ;
if (pts[s].size()>pts[t].size()) swap(pts[s], pts[t]);
for (auto it:pts[s]) pts[t].insert(it);
dsu[s]=t;
}
namespace force{
int sta[N], top, ans=INF, mini;
int tem[N], dsu2[N], usiz;
inline int find2(int p) {return dsu2[p]==p?p:dsu2[p]=find2(dsu2[p]);}
void solve() {
memset(vis, 0, sizeof(vis));
for (int i=1; i<=n; ++i)
for (int j=1; j<=m; ++j)
if (!vis[find(id[i][j])]) {
if (!s[i][j]) sta[top++]=find(id[i][j]);
else tem[++usiz]=find(id[i][j]);
vis[find(id[i][j])]=1;
}
// cout<<"---sta---"<<endl;
// for (int i=0; i<top; ++i) {for (auto it:pts[sta[i]]) cout<<"("<<it.fir<<','<<it.sec<<") "; cout<<endl;}
// cout<<"---tem---"<<endl;
// for (int i=1; i<=usiz; ++i) {for (auto it:pts[tem[i]]) cout<<"("<<it.fir<<','<<it.sec<<") "; cout<<endl;}
int lim=1<<top;
for (int s=0; s<lim; ++s) {
for (int i=1; i<=tot; ++i) dsu2[i]=i;
for (int i=0; i<top; ++i) if (s&(1<<i)) {
for (auto it:to[sta[i]]) {
int v=it.fir;
dsu2[find2(sta[i])]=find2(v);
}
}
int rot=find2(tem[1]);
// cout<<rot<<endl;
for (int i=2; i<=usiz; ++i) if (find2(tem[i])!=rot) goto jump;
if (__builtin_popcount(s)<ans) ans=__builtin_popcount(s), mini=s;
jump: ;
}
printf("%d\n", ans+(usiz!=0));
if (!(ans+(usiz!=0))) return ;
for (int i=0; i<top; ++i) if (mini&(1<<i))
for (auto it:pts[sta[i]]) s[it.fir][it.sec]=1;
puts("+");
for (int i=1; i<=n; ++i) {for (int j=1; j<=m; ++j) printf("%d", s[i][j]); printf("\n");}
for (int i=0; i<top; ++i) if (mini&(1<<i)) {
for (int j=1; j<=n; ++j) for (int k=1; k<=m; ++k) s[j][k]=0;
for (auto it:pts[sta[i]]) s[it.fir][it.sec]=1;
puts("-");
for (int i=1; i<=n; ++i) {for (int j=1; j<=m; ++j) printf("%d", s[i][j]); printf("\n");}
}
}
}
void print() {
int u=n, v=m;
if (rev) {
for (int i=1; i<=max(n, m); ++i)
for (int j=i+1; j<=max(n, m); ++j)
swap(s[i][j], s[j][i]);
swap(u, v);
for (int i=1; i<=u; ++i) {for (int j=1; j<=v; ++j) printf("%d", s[i][j]); printf("\n");}
}
else for (int i=1; i<=n; ++i) {for (int j=1; j<=m; ++j) printf("%d", s[i][j]); printf("\n");}
}
namespace task1{
int tem[N], usiz;
void solve() {
memset(vis, 0, sizeof(vis));
for (int i=1; i<=n; ++i)
for (int j=1; j<=m; ++j)
if (!vis[find(id[i][j])]) {
if (s[i][j]) tem[++usiz]=find(id[i][j]);
vis[find(id[i][j])]=1;
}
printf("%d\n", usiz);
for (int i=1; i<=usiz; ++i) {
for (int i=1; i<=n; ++i) for (int j=1; j<=m; ++j) s[i][j]=0;
for (auto it:pts[tem[i]]) s[it.fir][it.sec]=1;
puts("+");
print();
}
}
}
namespace task{
char bkp[N][N];
int black[N*N], white[N*N], bcnt, wcnt;
void solve() {
memset(vis, 0, sizeof(vis));
for (int i=1; i<=n; ++i)
for (int j=1; j<=m; ++j)
if (!vis[find(id[i][j])]) {
if (!s[i][j]) white[++wcnt]=find(id[i][j]);
else black[++bcnt]=find(id[i][j]);
vis[find(id[i][j])]=1;
}
if (!bcnt) {puts("0"); return ;}
if (bcnt==1) {
puts("1");
puts("+");
print();
return ;
}
for (int i=1; i<=wcnt; ++i) if (to[white[i]].size()==bcnt) {
puts("2");
for (auto it:pts[white[i]]) s[it.fir][it.sec]=1;
puts("+");
print();
for (int i=1; i<=n; ++i) for (int j=1; j<=m; ++j) s[i][j]=0;
puts("-");
for (auto it:pts[white[i]]) s[it.fir][it.sec]=1;
print();
return ;
}
puts("3");
for (int i=1; i<=n; ++i) for (int j=1; j<=m; ++j) bkp[i][j]=s[i][j];
puts("+");
for (int i=1; i<=n; ++i) for (int j=1; j<=m; ++j) s[i][j]=0;
for (int i=1; i<=n; ++i) s[i][1]=1;
for (int i=1; i<=n; i+=2) for (int j=1; j<m; ++j) s[i][j]=1;
for (int i=2; i<=n; i+=2) for (int j=2; j<m; ++j) if (bkp[i][j]) s[i][j]=1;
print();
puts("+");
for (int i=1; i<=n; ++i) for (int j=1; j<=m; ++j) s[i][j]=0;
for (int i=1; i<=n; ++i) s[i][m]=1;
for (int i=2; i<=n; i+=2) for (int j=2; j<=m; ++j) s[i][j]=1;
for (int i=1; i<=n; i+=2) for (int j=2; j<m; ++j) if (bkp[i][j]) s[i][j]=1;
print();
puts("-");
for (int i=1; i<=n; ++i) for (int j=1; j<=m; ++j) s[i][j]=1;
for (int i=1; i<=n; ++i) if (bkp[i][1]) s[i][1]=0;
for (int i=1; i<=n; ++i) if (bkp[i][m]) s[i][m]=0;
print();
return ;
}
}
signed main()
{
freopen("bitbit.in", "r", stdin);
freopen("bitbit.out", "w", stdout);
scanf("%d%d", &n, &m);
for (int i=1; i<=n; ++i) {
scanf("%s", s[i]+1);
for (int j=1; j<=m; ++j) s[i][j]-='0';
}
if (n>m) {
rev=1;
for (int i=1; i<=max(n, m); ++i)
for (int j=i+1; j<=max(n, m); ++j)
swap(s[i][j], s[j][i]);
swap(n, m);
}
for (int i=1; i<=n; ++i) for (int j=1; j<=m; ++j) id[i][j]=++tot, pts[tot].insert({i, j});
for (int i=1; i<=tot; ++i) dsu[i]=i;
for (int i=1; i<=n; ++i)
for (int j=1; j<=m; ++j)
for (int k=0; k<4; ++k) {
int x=i+dlt[k][0], y=j+dlt[k][1];
if (x>=1&&x<=n&&y>=1&&y<=m&&s[x][y]==s[i][j]) uni(id[i][j], id[x][y]);
}
for (int i=1; i<=n; ++i)
for (int j=1; j<=m; ++j)
for (int k=0; k<4; ++k) {
int x=i+dlt[k][0], y=j+dlt[k][1];
if (x>=1&&x<=n&&y>=1&&y<=m&&s[x][y]!=s[i][j]) to[find(id[i][j])][find(id[x][y])]=1;
}
for (int i=1; i<=n; ++i) for (int j=1; j<=m; ++j) if (s[i][j]&&!vis[find(id[i][j])]) ++cnt, vis[find(id[i][j])]=1;
// cout<<"cnt: "<<cnt<<endl;
// force::solve();
if (n==1) task1::solve();
else task::solve();
return 0;
}
