CF946D Timetable 动态规划

预处理出每一行去掉$k$个1能获得的最小代价

之后做一次分组背包$dp$即可

预处理可以选择暴力枚举区间...

复杂度$O(n^3)$

#include <set>
#include <map>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
namespace remoon {
    #define re register
    #define ri register int
    #define ll long long
    #define tpr template <typename ra>
    #define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++)
    #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --)
    #define gc getchar
    inline int read() {
        int p = , w = ; char c = gc();
        while(c > '' || c < '') { if(c == '-') w = -; c = gc(); }
        while(c >= '' && c <= '') p = p *  + c - '', c = gc();
        return p * w;
    }
    int wr[], rw;
    #define pc(iw) putchar(iw)
    tpr inline void write(ra o, char c = '\n') {
        if(!o) pc('');
        if(o < ) o = -o, pc('-');
        while(o) wr[++ rw] = o % , o /= ;
        while(rw) pc(wr[rw --] + '');
        pc(c);
    }
    tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; }
    tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; }
    tpr inline bool ckmin(ra &a, ra b) { return (a > b) ? a = b,  : ; }
    tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b,  : ; }
}
using namespace std;
using namespace remoon;

#define sid 505

int n, m, k;
char s[sid][sid];
int pre[sid], suf[sid];
int f[sid][sid], g[sid][sid];

inline void Pre() {
    rep(i, , n) {
        memset(pre, , sizeof(pre));
        memset(suf, , sizeof(suf));
        rep(l, , m) pre[l] = pre[l - ] + (s[i][l] == '');
        drep(l, m, ) suf[l] = suf[l + ] + (s[i][l] == '');
        if(pre[m] == ) continue;

        memset(f[i], , sizeof(f[i]));
        rep(l, , m) rep(r, l, m)
        if(s[i][l] == '' && s[i][r] == '')
        cmin(f[i][pre[l - ] + suf[r + ]], r - l + );
        rep(j, pre[m], k) f[i][j] = ;
    }
}

inline void DP() {
    memset(g, , sizeof(g));
    g[][] = ;
    rep(i, , n) {
        rep(i1, , k) rep(i2, , k)
        if(i1 + i2 <= k) cmin(g[i][i1 + i2], g[i - ][i1] + f[i][i2]);
    }
    int ans = 2e9;
    rep(i, , k) cmin(ans, g[n][i]);
    write(ans);
}

int main() {
    n = read(); m = read(); k = read();
    rep(i, , n) scanf("%s", s[i] + );
    Pre(); DP();
    return ;
}