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  • 数据结构:二维单调队列模板

    模板:
    二维单调队列维护二维区间最大值、最小值

    点击查看折叠代码块
    /*
    二维单调队列可以维护区间的最大值,最小值
    */
    #include <bits/stdc++.h>
    using namespace std;
    typedef long long int LL;
    const int MAXN = 1000 + 10;
    const int INF = 0x3f3f3f3f;
    int n, m, k, num[MAXN][MAXN], minNum[MAXN][MAXN], maxNum[MAXN][MAXN];
    void solve(int type, int seg[][MAXN]){
        deque<pair<int, int> > deq;
        for (int i = 1; i <= n; i++){ //求行子段的最值。
            deq.clear();
            for (int j = 1; j <= m; j++){
                while (!deq.empty() && j - deq.front().second >= k){ deq.pop_front(); }
                if (type){
                    while (!deq.empty() && deq.back().first < num[i][j]){ deq.pop_back(); }
                }
                else{
                    while (!deq.empty() && deq.back().first > num[i][j]){ deq.pop_back(); }
                }
                deq.push_back(make_pair(num[i][j], j));
                seg[i][j] = deq.front().first;
            }
        }
        for (int j = 1; j <= m; j++){ //求列的最值
            deq.clear();
            for (int i = 1; i <= n; i++){
                while (!deq.empty() && i - deq.front().second >= k){ deq.pop_front(); }
                if (type){
                    while (!deq.empty() && deq.back().first < seg[i][j]){ deq.pop_back(); }
                }
                else{
                    while (!deq.empty() && deq.back().first > seg[i][j]){ deq.pop_back(); }
                }
                deq.push_back(make_pair(seg[i][j], i));
                seg[i][j] = deq.front().first;
            }
        }
    }
    int main(){
        while (~scanf("%d%d%d", &n, &m, &k)){
            for (int i = 1; i <= n; i++){
                for (int j = 1; j <= m; j++){
                    scanf("%d", &num[i][j]);
                }
            }
            int ans = INF;
            solve(0, minNum); solve(1, maxNum);//求区间最小值,最大值
            for (int i = k; i <= n; i++){
                for (int j = k; j <= m; j++){
                    ans = min(ans, maxNum[i][j] - minNum[i][j]);
                }
            }
            printf("%d
    ", ans);
        }
        return 0;
    }
    
    你将不再是道具,而是成为人如其名的人
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  • 原文地址:https://www.cnblogs.com/wsl-lld/p/13393614.html
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