HDU - 1506 Largest Rectangle in a Histogram(dp)
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题意:已知n个高度不一、宽度相同的矩形并列排放,求所形成的图形中最大的矩形面积。
分析:
1、对于每一个矩形,分别算出它左边连续比它高的矩形中最左边的下标,右边同理。通过(r[i] - l[i] + 1) * a[i]比较得到最大的矩形面积。
2、将一组高度依次降低的矩形看成一个整体,如果该矩形比这个整体最矮的矩形都矮,长度可直接延伸过去,通过tmp = l[tmp - 1],依次向左比较,直到找到最左边的下标。
#include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define lowbit(x) (x & (-x)) const double eps = 1e-8; inline int dcmp(double a, double b){ if(fabs(a - b) < eps) return 0; return a > b ? 1 : -1; } typedef long long LL; typedef unsigned long long ULL; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const LL LL_INF = 0x3f3f3f3f3f3f3f3f; const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const int MAXN = 100000 + 10; const int MAXT = 10000 + 10; using namespace std; int a[MAXN], l[MAXN], r[MAXN]; int main(){ int n; while(scanf("%d", &n) == 1){ if(n == 0) return 0; for(int i = 0; i < n; ++i){ scanf("%d", &a[i]); } l[0] = 0; for(int i = 1; i < n; ++i){ int tmp = i; while(tmp > 0 && a[tmp - 1] >= a[i]) tmp = l[tmp - 1]; l[i] = tmp; } r[n - 1] = n - 1; for(int i = n - 2; i >= 0; --i){ int tmp = i; while(tmp < n - 1 && a[tmp + 1] >= a[i]) tmp = r[tmp + 1]; r[i] = tmp; } LL ans = 0; for(int i = 0; i < n; ++i){ ans = max(ans, (LL)(r[i] - l[i] + 1) * a[i]); } cout << ans << endl; } return 0; }
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