Codeforces 946D - Timetable 【动态规划】

D. Timetable
time limit per test 
2 seconds
memory limit per test     
256 megabytes

Ivan is astudent at Berland State University (BSU). There are
n
days in Berland week, and each of these days Ivan might have some classesat the university.

There are
m
working hours during each Berland day, and each lesson at the universitylasts exactly one hour. If at some day Ivan's first lesson is during
i-th
hour, and last lesson is during j-th
hour, then he spends j - i + 1
hours in the university during this day. If there are no lessons duringsome day, then Ivan stays at home and therefore spends
0
hours in the university.

Ivan doesn'tlike to spend a lot of time in the university, so he has decided to skip somelessons. He cannot skip more
than k
lessons during the week. After deciding which lessons he should skip andwhich he should attend, every day Ivan will enter the university right beforethe start of the first lesson he does not skip, and leave it after the end ofthe last lesson he decides to
attend. If Ivan skips all lessons during someday, he doesn't go to the university that day at all.

Given
n,
m,
k
and Ivan's timetable, can you determine the minimum number of hours hehas to spend in the university during one week, if he cannot skip more than
k
lessons?
Input

The first linecontains three integers
n,
m
and k
(1 ≤ n, m ≤ 500,
0 ≤ k ≤ 500)
— the numberof days in the Berland week, the number of working hours during each day, andthe number of lessons Ivan can skip, respectively.

Then
n
lines follow, i-th
linecontaining a binary string of m
characters. If j-th
character in i-th
line is 1,
then Ivan has a lesson on i-th
day during j-th
hour (if itis 0,
there is no such lesson).
Output
Print the minimum number of hours Ivanhas
to spend in the university during the week if he skips not more than k
lessons.
Examples
Input

2 5 1

01001

10110
Output

5
Input

2 5 0

01001

10110
Output

8
Note

In the firstexample Ivan can skip any of two lessons during the first day, so he spends
1
hour during the first day and 4
hours during the second day.
In the second example Ivan can't skipany lessons,
so he spends 4
hours every day
 

【题意】

现在告诉你一星期有n天，每天有m节课，这些课里面只有一部分需要上，每节课需要上一个小时，每天你需要在学校待的时间为你需要上的最后一节课结束的时间减去第一节课开始的时间。

你可以翘掉其中的k节课，问一个星期最少在学校里待多久。

【思路】

首先我们计算出对于每一天翘掉j节课要在学校里待的时间，即用num[i][j]表示第i天的课翘j门需要在校的最少时间。

然后用dp[i][j]表示第i天的课翘j门需要在校的最少时间，用前面计算出来的dp数组和num数组转移即可。

状态转移方程为

dp[i][j]=min{dp[i-1][j-o]+num[i][o]} (0<=0<=j)

#include <cstdio>
#include <iostream>
#include <cmath>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define rush() int T;scanf("%d",&T);while(T--)

typedef long long ll;
const int maxn = 505;
const ll mod = 1e9+7;
const ll INF = 1e18;
const double eps = 1e-9;

char s[maxn];
int num[maxn][maxn];
int dp[maxn][maxn];
vector<int>vec[maxn];

int main()
{
int n,m,k;
scanf("%d%d%d",&n,&m,&k);
for(int i=1;i<=n;i++)
{
scanf("%s",s+1);
int len=strlen(s+1);
for(int j=1;j<=len;j++)
{
if(s[j]=='1') vec[i].push_back(j);
}
}
for(int i=1;i<=n;i++)
{
int tmp=vec[i].size();
if(tmp==0) continue;
for(int len=1;len<=tmp;len++)
{
int Min=vec[i][len-1]-vec[i][0]+1;
for(int j=0;j+len-1<tmp;j++)
{
Min=min(Min,vec[i][j+len-1]-vec[i][j]+1);
}
num[i][tmp-len]=Min;
}
}
for(int i=0;i<=k;i++) dp[1][i]=num[1][i];
for(int i=2;i<=n;i++)
{
for(int j=0;j<=k;j++)
{
int Min=dp[i-1][j]+num[i][0];
for(int o=0;o<=j;o++)
{
Min=min(Min,dp[i-1][j-o]+num[i][o]);
}
dp[i][j]=Min;
}
}
printf("%d\n",dp
[k]);
}