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 = 0, w = 1; char c = gc();
        while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }
        while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
        return p * w;
    }
    int wr[50], rw;
    #define pc(iw) putchar(iw)
    tpr inline void write(ra o, char c = '
') {
        if(!o) pc('0');
        if(o < 0) o = -o, pc('-');
        while(o) wr[++ rw] = o % 10, o /= 10;
        while(rw) pc(wr[rw --] + '0');
        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, 1 : 0; }
    tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b, 1 : 0; }
}
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, 1, n) {
        memset(pre, 0, sizeof(pre));
        memset(suf, 0, sizeof(suf));
        rep(l, 1, m) pre[l] = pre[l - 1] + (s[i][l] == '1');
        drep(l, m, 1) suf[l] = suf[l + 1] + (s[i][l] == '1');
        if(pre[m] == 0) continue;
        
        memset(f[i], 56, sizeof(f[i]));
        rep(l, 1, m) rep(r, l, m)
        if(s[i][l] == '1' && s[i][r] == '1')
        cmin(f[i][pre[l - 1] + suf[r + 1]], r - l + 1);
        rep(j, pre[m], k) f[i][j] = 0;
    }
}

inline void DP() {
    memset(g, 56, sizeof(g));
    g[0][0] = 0;
    rep(i, 1, n) {
        rep(i1, 0, k) rep(i2, 0, k)
        if(i1 + i2 <= k) cmin(g[i][i1 + i2], g[i - 1][i1] + f[i][i2]);
    }
    int ans = 2e9;
    rep(i, 0, k) cmin(ans, g[n][i]);
    write(ans);
}

int main() {
    n = read(); m = read(); k = read();
    rep(i, 1, n) scanf("%s", s[i] + 1);
    Pre(); DP();
    return 0;
}