[Codeforces 946D]Timetable

Description

题库链接

给你一个 \(N\times M\) 的 \(01\) 矩阵，你可以从中将一些 \(1\) 变为 \(0\) ，最多 \(K\) 次。使操作之后使得每行最远的 \(1\) 间距和最小。输出最小值。

\(1\leq N,M,K\leq 500\)

Solution

显然可以预处理一个数组 \(a_{i,j}\) 表示第 \(i\) 行操作 \(j\) 次后最小的间距。这个可以用 \(O(N^3)\) 枚举出来的。

其次记 \(f_{i,j}\) 表示前 \(i\) 行操作 \(j\) 次后，最小间距和，也是 \(O(N^3)\) 的 \(DP\) 。

最后答案就是 \(\min\limits_{i=1}^k f_{n,i}\) 。

Code

//It is made by Awson on 2018.3.10
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 500, INF = ~0u>>1;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, m, k, a[N+5][N+5], b[N+5], top, len, l[N+5];
char ch[N+5];
int f[N+5][N+5];

void get_a(int id) {
    a[id][top] = 0;
    for (int i = 1; i <= top; i++)
    for (int j = 1; j+i-1 <= top; j++)
        a[id][top-i] = Min(a[id][top-i], b[j+i-1]-b[j]+1);
}
void work() {
    read(n), read(m), read(k);
    memset(a, 127/3, sizeof(a)); memset(f, 127/3, sizeof(f));
    for (int i = 1; i <= n; i++) {
    scanf("%s", ch); top = 0, len = strlen(ch);
    for (int j = 0; j < len; j++) if (ch[j] == '1') b[++top] = j;
    get_a(i); l[i] = top;
    }
    f[0][0] = 0;
    for (int i = 1; i <= n; i++)
    for (int j = 0; j <= k; j++)
        for (int q = 0; q <= j; q++)
        f[i][j] = Min(f[i][j], f[i-1][q]+a[i][j-q]);
    int ans = INF; for (int i = 0; i <= k; i++) ans = Min(ans, f[n][i]);
    printf("%d\n", ans);
}
int main() {
    work(); return 0;
}