/* 2013暑期多校联合训练——80+高校，300+队伍，10000元奖金，敬请期待~ Billboard Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6979 Accepted Submission(s): 3138 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 Author hhanger@zju Source HDOJ 2009 Summer Exercise（5） Recommend lcy 题意 题目意思是给定一个高和宽的广告排， 然后有N个人想在这上面张贴自己的广告， 每个题提供的广告都是高为1，宽由他们定制的广告， 求在满足先按行递增以及尽量满足左边空间的张贴方式， 输出广告所在行。如果不能张贴的话，输出-1。 */ /*在原来的线段树基础上完成了，再看看大神的，自己的又麻烦了*/ #include <iostream> #include<algorithm> #include<queue> #include<cmath> #include<string.h> #include<stdio.h> #include<stdlib.h> using namespace std; //#define maxn 50100 const int maxn = 200100; #define lson l , m , rt<<1 #define rson m+1 , r , rt<<1|1 int Max[maxn<<2]; int h,w,tag,ans; void PushUp(int rt) { Max[rt]=max(Max[rt<<1],Max[rt<<1|1]); } void build(int l,int r,int rt) { if(l==r) { Max[rt]=w; //printf("l = %d r = %d rt = %d Max[%d] = %d ",l,r,rt,rt,Max[rt]); return ; } int m = (l+r)>>1; build(lson); build(rson); PushUp(rt); //printf("l = %d r = %d rt = %d Max[%d] = %d ",l,r,rt,rt,Max[rt]); } void update(int p ,int dec, int l ,int r,int rt) { if(l==r) { Max[rt]-=dec; return ; } int m = (l+r)>>1; if(p<=m) update(p,dec,lson); else update(p,dec,rson); PushUp(rt); } int query(int sr,int l,int r,int rt) { if(tag==true) return 0; int m = (l+r)>>1; if(Max[rt]>=sr) { if(l==r) { //printf("%d ",l); ans=l; tag=true; return 0; } query(sr,lson); query(sr,rson); } } int main() { int sr,wi; while(~scanf("%d%d",&h,&w)) { scanf("%d",&wi); if (h > wi) h = wi; //这个是关键啊，没有的话就是“数组越界” //主要是题目坑啊，明明是200000，拿10^9来吓我们。 build(1,h,1); while(wi--) { scanf("%d",&sr); tag=false; if(sr>Max[1]) { printf("-1 "); } else { query(sr,1,h,1); printf("%d ",ans); update(ans,sr,1,h,1); // printf("%d ",Max[4]); } } } return 0; }
//NotOnlySuccess大神的 #include <cstdio> #include <algorithm> using namespace std; #define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 const int maxn = 222222; int h , w , n; int MAX[maxn<<2]; void PushUP(int rt) { MAX[rt] = max(MAX[rt<<1] , MAX[rt<<1|1]); } void build(int l,int r,int rt) { MAX[rt] = w; if (l == r) return ; int m = (l + r) >> 1; build(lson); build(rson); } int query(int x,int l,int r,int rt) { if (l == r) { MAX[rt] -= x; return l; } int m = (l + r) >> 1; int ret = (MAX[rt<<1] >= x) ? query(x , lson) : query(x , rson); PushUP(rt); return ret; } int main() { while (~scanf("%d%d%d",&h,&w,&n)) { if (h > n) h = n; build(1 , h , 1); while (n --) { int x; scanf("%d",&x); if (MAX[1] < x) puts("-1"); else printf("%d ",query(x , 1 , h , 1)); } } return 0; }