POJ2559/HDU1506 Largest Rectangle in a Histogram (cartesian tree)

Die datenstruktur ist erataunlich!

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long
#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")

#else

#define D_e_Line ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;

const int N = 100007;

int n;

struct node{
    int val, pri, fa, l, r;

    bool operator < (const node &com)const{
		return val < com.val;
	}
}t[N];

int root;
long long ans;

inline int Build(){
	R(i,1,n){
		int k;
		for(k = i - 1; t[k].pri > t[i].pri; k = t[k].fa);
		t[i].l = t[k].r;
		t[k].r = i;
		t[i].fa = k;
		t[t[i].l].fa = i;
	}
	return t[0].r;
}

inline long long DFS(int x){
	if(!x) return 0;
	long long width = DFS(t[x].l) + DFS(t[x].r) + 1;
	long long tmp = t[x].pri * width;
	ans = Max(ans, tmp);
	return width;
}
int main(){
    while(scanf("%d", &n) && n){
    	t[0].l = t[0].r = t[0].fa = t[0].pri = t[0].val = 0;
        R(i,1,n){
        	t[i].val = i;
            io >> t[i].pri;
            t[i].l = t[i].r = t[i].fa = 0;
        }

        root = Build();

        ans = 0;

        DFS(root);

        printf("%lld\n", ans);
    }
    return 0;
}