POJ2082 最大矩形面积 单调栈

【题目链接】

          点击打开链接

【题意】

          给了一些h*w的矩形，大小可不等，求可组成的最大矩形面积是多少？单调栈经典题，做法和点击打开链接 完全相同。

【AC代码】

//
//Created by just_sort 2016/1/5
//Copyright (c) 2016 just_sort.All Rights Reserved
//

//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/hash_policy.hpp>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <time.h>
#include <cstdlib>
#include <cstring>
#include <sstream> //isstringstream
#include <iostream>
#include <algorithm>
using namespace std;
//using namespace __gnu_pbds;
typedef long long LL;
typedef pair<int, LL> pp;
#define REP1(i, a, b) for(int i = a; i < b; i++)
#define REP2(i, a, b) for(int i = a; i <= b; i++)
#define REP3(i, a, b) for(int i = a; i >= b; i--)
#define CLR(a, b)     memset(a, b, sizeof(a))
#define MP(x, y)      make_pair(x,y)
const int maxn = 1e6 + 10;
const int maxm = 2e5;
const int maxs = 10;
const int maxp = 1e3 + 10;
const int INF  = 1e9;
const int UNF  = -1e9;
const int mod  = 1e9 + 7;
int gcd(int x, int y) {return y == 0 ? x : gcd(y, x % y);}
//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;
//head
struct node{
    int height, width;
}s[maxn];
int top;
int getmax(int a, int b)
{
    return a > b ? a : b;
}
int main()
{
    int n, height, width, tmp, ans, tot, top;
    while(scanf("%d", &n) != EOF && n != -1)
    {
        top = ans = 0;
        REP1(i, 0, n){
            scanf("%d%d", &width, &height);
            tmp = 0;
            while(top > 0 && s[top - 1].height >= height)
            {
                tot = s[top - 1].height * (s[top - 1].width + tmp);
                ans = getmax(ans, tot);
                tmp += s[top - 1].width;
                --top;
            }
            s[top].height = height;
            s[top].width = width + tmp;
            top++;
        }
        tmp = 0;
        while(top > 0)
        {
            tot = s[top - 1].height * (s[top - 1].width + tmp);
            ans = getmax(ans, tot);
            tmp += s[top - 1].width;
            --top;
        }
        cout << ans << endl;
    }
    return 0;
}