題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=2795

題面：

Billboard Time Limit: 20000/8000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16167????Accepted Submission(s): 6837


Problem Description At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is the place where all possible announcements are posted: nearest programming competitions, changes in the dining room menu, and other important information.

On September 1, the billboard was empty. One by one, the announcements started being put on the billboard.

Each announcement is a stripe of paper of unit height. More specifically, the i-th announcement is a rectangle of size 1 * wi.

When someone puts a new announcement on the billboard, she would always choose the topmost possible position for the announcement. Among all possible topmost positions she would always choose the leftmost one.

If there is no valid location for a new announcement, it is not put on the billboard (that's why some programming contests have no participants from this university).

Given the sizes of the billboard and the announcements, your task is to find the numbers of rows in which the announcements are placed. ?
Input There are multiple cases (no more than 40 cases).

The first line of the input file contains three integer numbers, h, w, and n (1 <= h,w <= 10^9; 1 <= n <= 200,000) - the dimensions of the billboard and the number of announcements.

Each of the next n lines contains an integer number wi (1 <= wi <= 10^9) - the width of i-th announcement. ?
Output For each announcement (in the order they are given in the input file) output one number - the number of the row in which this announcement is placed. Rows are numbered from 1 to h, starting with the top row. If an announcement can't be put on the billboard, output "-1" for this announcement. ?
Sample Input

3 5 5 2 4 3 3 3 ?
Sample Output

1 2 1 3 -1

題目大意：

? ? 給一個H*M的板，每次貼1*w的海報，要求貼在能貼的行數(shù)最小的行的最左邊，且不能覆蓋，若不能貼，則輸出-1，否則輸出該行行號。

解題：

? ? 操作是很簡單的線段樹，但不得不說，這題考的是想法，我自己是想不到用線段樹去維護(hù)區(qū)間行的最大值。線段樹熟悉之后，更多得考的是想法和如何靈活地應(yīng)用線段樹。

代碼：

#include 
#include 
#include 
#define maxn 200010
using namespace std;
int maxw[maxn<<2];
int h,w,n,ans,tw;
int max(int a,int b)
{
    return a>b?a:b;
}
//建樹 
void build(int le,int ri,int u)
{
   //初值都賦為w 
   maxw[u]=w;
   if(le==ri)return;
   int mid=(le+ri)>>1;
   build(le,mid,u<<1);
   build(mid+1,ri,(u<<1)+1);
}

int query(int le,int ri,int u,int val)
{
    int res;
    if(le==ri)
    {
    	//更新 
        maxw[u]-=val;
        return le;
    }
    //如果左邊可以放下，就放左邊，即滿足盡量放上邊 
    if(maxw[u<<1]>=val)
       res=query(le,(le+ri)>>1,u<<1,val);
    //左邊放不下，放右邊 
    else
       res=query(((le+ri)>>1)+1,ri,(u<<1)+1,val);
    //更新當(dāng)前節(jié)點最大值 
    maxw[u]=max(maxw[u<<1],maxw[(u<<1)+1]);
    return res;
}
int main()
{
    while(~scanf("%d%d%d",&h,&w,&n))
    {
        if(h>n)
          h=n;
        build(1,h,1);
        for(int i=0;i