• 2019牛客暑期多校训练营 第二场


     A.Eddy Walker

    B.Eddy Walker 2

     C.Go on Strike!

     D.Kth Minimum Clique

     E.MAZE

    F.Partition problem

     G.Polygons

     H.Second Large Rectangle

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 const int N=2000;
     4 int n,m,k,l[N],r[N],h[N],a[N*N],ans,kk;
     5 char s[N][N];
     6 int main() {
     7     scanf("%d%d", &n, &m);
     8     for (int j=0;j<m;j++){
     9             l[j]=-1;
    10             r[j]=m;
    11     }
    12     set< pair<pair<int, int>, int> > ss;
    13     for (int i = 0; i < n; i++) {
    14         scanf("%s", &s[i]);
    15         k = -1;
    16         for (int j = 0; j < m; j++) {
    17             l[j] = max(k, l[j]);
    18             if (s[i][j] == '1') {
    19                 h[j]++;
    20             } else {
    21                 h[j] = 0;
    22                 l[j] = -1;
    23                 k = j;
    24             }
    25         }
    26         k = m;
    27         for (int j = m - 1; j >= 0; j--) {
    28             r[j] = min(r[j], k);
    29             if (s[i][j] == '0') {
    30                 k = j;
    31                 r[j] = m;
    32             }
    33         }
    34         ss.clear();
    35         for (int j = 0; j < m; ++j) {
    36             ss.insert(make_pair(make_pair(l[j], r[j]), h[j]));
    37         }
    38         for (auto item : ss) {
    39             a[++kk] = item.second * (item.first.second - item.first.first - 1);
    40             a[++kk] = item.second * (item.first.second - item.first.first - 2);
    41             a[++kk] = (item.second - 1) * (item.first.second - item.first.first - 1);
    42         }
    43     }
    44     sort(a+1, a+kk+1 ,greater<int>());
    45     printf("%d
    ",max(0,a[2]));
    46 }
    View Code

    I.Inside A Rectangle

     J.Subarray

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  • 原文地址:https://www.cnblogs.com/Accpted/p/11218501.html
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