Day8 - C - Largest Rectangle in a Histogram HDU - 1506

A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left shows the histogram that consists of rectangles with the heights 2, 1, 4, 5, 1, 3, 3, measured in units where 1 is the width of the rectangles:

Usually, histograms are used to represent discrete distributions, e.g., the frequencies of characters in texts. Note that the order of the rectangles, i.e., their heights, is important. Calculate the area of the largest rectangle in a histogram that is aligned at the common base line, too. The figure on the right shows the largest aligned rectangle for the depicted histogram.

InputThe input contains several test cases. Each test case describes a histogram and starts with an integer n, denoting the number of rectangles it is composed of. You may assume that 1 <= n <= 100000. Then follow n integers h1, ..., hn, where 0 <= hi <= 1000000000. These numbers denote the heights of the rectangles of the histogram in left-to-right order. The width of each rectangle is 1. A zero follows the input for the last test case.OutputFor each test case output on a single line the area of the largest rectangle in the specified histogram. Remember that this rectangle must be aligned at the common base line.Sample Input

7 2 1 4 5 1 3 3
4 1000 1000 1000 1000
0

Sample Output

. 8 
4000 

idea: Monotone board title queue, attention monotonicity When dequeuing statistics length, will provide a set of data 74562333 (14 instead of 12 new length becomes longer enqueued

typedef long long LL;
typedef pair<LL, LL> PLL;

const int maxm = 1e5+5; 

int buf[maxm], q[maxm], len[maxm];

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    int n;
    while(cin >> n && n) {
        LL ans = 0;
        for(int i = 1; i <= n; ++i) {
            cin >> buf[i];
            len[i] = 1;
        }
        buf[n+1] = 0;
        int l = 0, r = -1, cnt;
        bool flag = false;
        for(int i = 1; i <= n+1; ++i) {
            cnt = 0; flag = false;
            while(l <= r && buf[q[r]] > buf[i]) {
                flag = true;
                cnt += len[q[r]];
                ans = max(ans, 1LL*cnt*buf[q[r]]);
                r--;
            }
            q[++r] = i;
            if(flag) len[i] = cnt+1;
        }
        cout << ans << "\n";
    }

    return 0;
}

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Cartesian tree can also be used to do, where attention is not strictly Cartesian tree, because there is a node of the same weight, length a is the number of points of his own descent + (1), using the memory of the search , or will T

typedef long long LL;
typedef pair<LL, LL> PLL;

const int maxm = 1e5+5; 

int buf[maxm], q[maxm], Left[maxm], Right[maxm], length[maxm];

int getlen(int u) {
    if(length[u]) return length[u];
    int len = 1;
    if(Left[u]) len += getlen(Left[u]);
    if(Right[u]) len += getlen(Right[u]);
    return length[u] = len;
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    int n;
    while(cin >> n && n) {
        LL ans = 0;
        for(int i = 1; i <= n; ++i) Left[i] = Right[i] = length[i] = 0;
        for(int i = 1; i <= n; ++i)
            cin >> buf[i];
        int top = 0;
        for(int i = 1; i <= n; ++i) {
            while(top && buf[q[top]] > buf[i]) {
                Left[i] = q[top];
                top--;
            }
            if(top)
                Right[q[top]] = i;
            q[++top] = i;
        }
        for(int i = 1; i <= n; ++i) {
            int len = 1;
            len = getlen(i);
            ans = max(ans, 1LL*len*buf[i]);
        }
        cout << ans << "\n";
    }

    return 0;
}

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