题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2795
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.
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.
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
题意：
一个广告牌，要在个广告牌上写广告，注意广告不能写在两行，只能写在一行，然后告诉你这个矩形的高度和宽度，高度意味着有多少行，一行占据一个高度，最后在给你要写的广告的数量，往这块广告牌上弄的原则就是尽量往上面的位置放，同一行尽量往左放。
思路：
把[1，h]当做一个区间，每个叶子结点的初始大小为 W ，利用线段树求解。每当某一行满足条件，就减去它相应的宽度。非叶子结点存放的对应区间的可用的最大宽度，所以判断能否存放时只需要和根节点比较即可。
1 #include <map> 2 #include <stack> 3 #include <queue> 4 #include <cmath> 5 #include <string> 6 #include <limits> 7 #include <cstdio> 8 #include <cstdlib> 9 #include <cstring> 10 #include <iostream> 11 #include <algorithm> 12 #define Scc(c) scanf("%c",&c) 13 #define Scs(s) scanf("%s",s) 14 #define Sci(x) scanf("%d",&x) 15 #define Sci2(x, y) scanf("%d%d",&x,&y) 16 #define Sci3(x, y, z) scanf("%d%d%d",&x,&y,&z) 17 #define Scl(x) scanf("%I64d",&x) 18 #define Scl2(x, y) scanf("%I64d%I64d",&x,&y) 19 #define Scl3(x, y, z) scanf("%I64d%I64d%I64d",&x,&y,&z) 20 #define Pri(x) printf("%d ",x) 21 #define Prl(x) printf("%I64d ",x) 22 #define Prc(c) printf("%c ",c) 23 #define Prs(s) printf("%s ",s) 24 #define For(i,x,y) for(int i=x;i<y;i++) 25 #define For_(i,x,y) for(int i=x;i<=y;i++) 26 #define FFor(i,x,y) for(int i=x;i>y;i--) 27 #define FFor_(i,x,y) for(int i=x;i>=y;i--) 28 #define Mem(f, x) memset(f,x,sizeof(f)) 29 #define LL long long 30 #define ULL unsigned long long 31 #define MAXSIZE 200005 32 #define INF 0x3f3f3f3f 33 const int mod=1e9+7; 34 const double PI = acos(-1.0); 35 36 using namespace std; 37 struct node 38 { 39 int left,right; 40 int maxx; 41 } tree[MAXSIZE*4]; 42 int a[MAXSIZE]; 43 int h,w; 44 void build (int i, int left,int right) 45 { 46 tree[i].left=left; 47 tree[i].right=right; 48 if(left==right) 49 { 50 tree[i].maxx=w; 51 return ; 52 } 53 build(i<<1,left,(left+right)/2 ); 54 build(i<<1|1,(left+right)/2+1,right); 55 tree[i].maxx=max(tree[i<<1].maxx,tree[i<<1|1].maxx); 56 } 57 int query(int i,int l,int r,int v) 58 { 59 if(l==r) 60 { 61 tree[i].maxx-=v; 62 return l;// 返回叶子的位置，即行位置 63 } 64 int mid=(l+r)/2; 65 int sum=0; 66 if(v<=tree[i<<1].maxx) 67 sum=query(i<<1,l,mid,v); 68 else 69 sum= query(i<<1|1,mid+1,r,v); 70 tree[i].maxx= max(tree[i<<1].maxx,tree[i<<1|1].maxx); 71 return sum; 72 } 73 int main() 74 { 75 int n; 76 while(~Sci3(h,w,n)) 77 { 78 h=min(h,n);//这里是缩小建树时的储存空间，当h>n时，最多只需要n行即可。 79 build(1,1,h); 80 int a; 81 For_(i,1,n) 82 { 83 Sci(a); 84 if(tree[1].maxx<a)//判断是否还能放广告 85 Pri(-1); 86 else 87 Pri(query(1,1,h,a)); 88 } 89 } 90 return 0; 91 }