[题解]CF514D R2D2 and Droid Army

思路

首先，可以转化题意，找到一个极长的区间 \([l,r]\) 使得（其中 \(mx_i\) 表示 \([l,r]\) 区间中属性 \(i\) 的最大值）：

\[ \sum_{i = 1}^{m}mx_i \leq k \]

显然对于这个东西当 \(l,r\) 发生移动时，是极其好维护的，所以想到双指针。

因为 \(m \leq 5\)，所以我们可以直接开 \(m\) 个 ST 表维护即可。

Code

#include <bits/stdc++.h>  
#define re register  
#define int long long  
#define pot(x) (1 << x)  
  
using namespace std;  
  
const int N = 1e5 + 10,M = 10,K = 20;  
int n,m,k,ans;  
int lg[N],a[N];  
int arr[N][M];  
  
inline int read(){  
    int r = 0,w = 1;  
    char c = getchar();  
    while (c < '0' || c > '9'){  
        if (c == '-') w = -1;  
        c = getchar();  
    }  
    while (c >= '0' && c <= '9'){  
        r = (r << 3) + (r << 1) + (c ^ 48);  
        c = getchar();  
    }  
    return r * w;  
}  
  
inline void init(){  
    for (re int i = 2;i <= n;i++) lg[i] = lg[i / 2] + 1;  
}  
  
struct ST{  
    int dp[N][K];  
  
    inline void init(){  
        for (re int j = 1;j <= lg[n];j++){  
            for (re int i = 1;i + pot(j) - 1 <= n;i++) dp[i][j] = max(dp[i][j - 1],dp[i + pot(j - 1)][j - 1]);  
        }  
    }  
  
    inline int query(int l,int r){  
        int k = lg[r - l + 1];  
        return max(dp[l][k],dp[r - pot(k) + 1][k]);  
    }  
}st[M];  
  
inline bool check(int l,int r){  
    int sum = 0;  
    for (re int i = 1;i <= m;i++) sum += st[i].query(l,r);  
    return (sum <= k);  
}  
  
signed main(){  
    n = read();  
    m = read();  
    k = read();  
    init();  
    for (re int i = 1;i <= n;i++){  
        for (re int j = 1;j <= m;j++) arr[i][j] = read();  
    }  
    for (re int j = 1;j <= m;j++){  
        for (re int i = 1;i <= n;i++) st[j].dp[i][0] = arr[i][j];  
        st[j].init();  
    }  
    for (re int l = 1,r = 1;r <= n;r++){   
        while (l <= r && !check(l,r)) l++;  
        if (ans < r - l + 1){  
            ans = r - l + 1;  
            for (re int i = 1;i <= m;i++) a[i] = st[i].query(l,r);  
        }  
    }  
    for (re int i = 1;i <= m;i++) printf("%lld ",a[i]);  
    return 0;  
}