HDU 2795 Billboard

题目链接：https://vjudge.net/problem/HDU-2795

题目大意：

　　有一块长度为 H 宽度为 W 的广告牌，现在有 n 则广告要依次贴上去，第 i 则广告的长度为 1 宽度为 w[i]，广告要尽量贴得高，其次尽量靠左，求每则广告位于第几行，贴不上就输出-1。

分析：

　　如果广告牌的所有行剩余宽度的最大值大于等于 w[i]，那么这个广告肯定能贴在某一行，可以通过二分的方法继续判断以缩小区间，没错，就是最值线段树。

代码如下：

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
 
#define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef set< int > SI;
typedef vector< int > VI;
typedef map< int, int > MII;
typedef vector< LL > VL;
typedef vector< VL > VVL;
const double EPS = 1e-10;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 2e5 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

LL H, W, n; 
LL w[maxN];

#define lson l , mid , rt << 1 
#define rson mid + 1 , r , rt << 1 | 1

struct SegmentTree{
    int st[maxN << 2];
    
    inline void pushUp(int rt) {
        st[rt] = max(st[rt << 1], st[rt << 1 | 1]);
    }
    
    inline void pushDown(int rt) { }
    
    inline void build(int l, int r, int rt) {
        if(l >= r) {
            st[rt] = W;
            return;
        }
        int mid = (l + r) >> 1;
        build(lson);
        build(rson);
        pushUp(rt);
    }
    
    // 查到某一行能容纳 x 就更新 
    inline LL queryAndUpdate(LL x, int l, int r, int rt) {
        if(st[rt] < x) return -1;
        if(l >= r) {
            st[rt] -= x;
            return r;
        }
        LL ret = 0;
        int mid = (l + r) >> 1;
        if(st[rt << 1] >= x) ret = queryAndUpdate(x, lson); // 前面的行优先 
        else if(st[rt << 1 | 1] >= x) ret = queryAndUpdate(x, rson);
        
        pushUp(rt);
        return ret;
    }
};
SegmentTree segTr;

int main(){
    INIT();
    while(cin >> H >> W >> n) {
        For(i, 1, n) cin >> w[i];
        if(H >= n) H = n; // 行数大于 n 时，只有前 n 行有用 
        segTr.build(1, H, 1);
        For(i, 1, n) cout << segTr.queryAndUpdate(w[i], 1, H, 1) << endl;
    }
    return 0;
}