没看到每个格子只能染一次色。。？11
于是 \(f_{i,j}\) 预处理每块木板涂 \(j\) 次色最多有多少个涂正确的，这个直接暴力枚举最后一次涂色的区间就好了。有了 \(f\) 就是简单背包了。
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
#define fin(x) freopen(#x".in","r",stdin)
#define fout(x) freopen(#x".out","w",stdout)
#define fr(x) fin(x),fout(x);
#define Fr(x,y) fin(x),fout(y)
#define INPUT(_1,_2,FILE,...) FILE
#define IO(...) INPUT(__VA_ARGS__,Fr,fr)(__VA_ARGS__)
using namespace std;
using namespace __gnu_pbds;
#define mp make_pair
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cfast ios::sync_with_stdio(false);cin.tie(0),cout.tie(0)
#define ll long long
#define ull unsigned long long
#define intz(x,y) memset((x),(y),sizeof((x)))
char *p1,*p2,buf[100000];
#define nc() (p1==p2 && (p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
#define tup(x) array<int,(x)>
inline ll read(){ll x=0,f=1;char ch=nc();while(ch<48||ch>57){if(ch=='-')f=-1;ch=nc();}while(ch>=48&&ch<=57)x=x*10+ch-48,ch=nc();return x*f;
}
//void write(int x){cout<<x<<' ';}
//void write(pii x){cout<<"P("<<x.fi<<','<<x.se<<")\n";}
//void write(vector<auto>x){for(auto i:x)write(i);cout<<'\n';}
//void write(auto *a,int l,int r){for(int i=l;i<=r;i++)write(a[i]);cout<<'\n';}
inline ll lowbit(ll x){return x&-x;}
inline int pcount(ll x){for(int i=0,res=0;;res+=(x>>i)&1,i++)if(i>60)return res;
}//struct mt{
// ll v;
// mt(){v=0;}
// mt(int x){this->v=x;}
// inline mt operator+(mt x){return {(v+x.v)%mod};}
// inline mt operator-(mt x){return {(v+mod-x.v)%mod};}
// inline mt operator*(mt x){return {1ll*v*x.v%mod};}
//};
//inline void add(mt &x,mt y){x=x+y;}
//mt qp(mt x,int y){mt res(1);for(;y;x=x*x,y>>=1)if(y&1)res=res*x;return res;}
const int N=55,M=2505;
#define int ll
char s[N][N];int f[N][N][M],c[N][2],g[M],tmp[M];
inline void UesugiErii(){int n,m,t;cin>>n>>m>>t;for(int i=1;i<=n;i++)cin>>(s[i]+1);for(int i=1;i<=n;i++){for(int j=1;j<=m;j++)c[j][0]=c[j-1][0],c[j][1]=c[j-1][1],++c[j][s[i][j]-'0'];for(int j=1;j<=m;j++)for(int p=1;p<=t;p++){f[i][j][p]=f[i][j-1][p];for(int k=0;k<j;k++)f[i][j][p]=max(f[i][j][p],f[i][k][p-1]+max(c[j][0]-c[k][0],c[j][1]-c[k][1]));}}for(int i=1;i<=n;i++){for(int j=1;j<=t;j++)tmp[j]=g[j];for(int k=1;k<=t;k++)for(int j=t;j>=k;j--)tmp[j]=max(tmp[j],g[j-k]+f[i][m][k]);swap(tmp,g);}cout<<g[t];
}
signed main(){//IO();//cfast;int _=1;//cin>>_;for(;_;_--)UesugiErii();return 0;
}