传送门
文章目录
- 题意:
- 思路:
题意:
n,m,k≤500n,m,k\le500n,m,k≤500
思路:
将其转换成背包的模型,就可以想出来一个很明显的dpdpdp状态:f[i][j]f[i][j]f[i][j]表示前iii行花费了jjj的最小代价,我们只需要预处理出来g[i][j]g[i][j]g[i][j]表示第iii行花费了jjj之后这一行的最小代价,就可以通过枚举jjj和ttt按照如下转移:f[i][j]=max(f[i][j],f[i−1][t]+g[i][j−t])f[i][j]=max(f[i][j],f[i-1][t]+g[i][j-t])f[i][j]=max(f[i][j],f[i−1][t]+g[i][j−t])
现在我们搞一下g[i][j]g[i][j]g[i][j]就好啦,可以发现我们每一行去除的111一定是从头或者从尾去除,那么再预处理两个数组l[i][j],r[i][j]l[i][j],r[i][j]l[i][j],r[i][j]表示第iii行从左删jjj个(从右删jjj个)之后的左边(右边)111的位置,这个很容易处理出来,之后就可以枚举去除的个数,再枚举从头去除的个数,直接O(1)O(1)O(1)转移即可。
复杂度O(n3)O(n^3)O(n3),cfcfcf测评机器很快,所以能过。
// Problem: D. Timetable
// Contest: Codeforces - Educational Codeforces Round 39 (Rated for Div. 2)
// URL: https://codeforces.com/contest/946/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=510,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n,m,k;
int f[N][N];//前i行,用了j个操作的最小值
int l[N][N],r[N][N],g[N][N];
char s[N];int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);cin>>n>>m>>k;memset(g,0x3f,sizeof(g));memset(f,0x3f,sizeof(f));for(int i=0;i<N;i++) f[0][i]=0;for(int i=1;i<=n;i++) {scanf("%s",s+1);int cnt=0;for(int j=1;j<=m;j++) {if(s[j]!='1') continue;l[i][cnt]=j; cnt++;}cnt=0;for(int j=m;j>=1;j--) {if(s[j]!='1') continue;r[i][cnt]=j; cnt++;}for(int j=0;j<=cnt;j++) {for(int k=0;k<=j;k++) {int t=j-k;g[i][j]=min(g[i][j],r[i][t]-l[i][k]+1);}}g[i][cnt]=0;for(int j=0;j<=k;j++) {//枚举前面的花费for(int now=0;now<=cnt;now++) {//枚举当前的花费if(j+now<=k) {//cout<<j<<' '<<now<<endl;f[i][j+now]=min(f[i][j+now],f[i-1][j]+g[i][now]);//cout<<f[i][j+now]<<' '<<f[i-1][k]<<' '<<g[i][now]<<' '<<i-1<<' '<<k<<endl;}}}}int ans=INF;for(int i=0;i<N;i++) ans=min(ans,f[n][i]);cout<<ans<<endl;return 0;
}
/**/