HDU 4614 Vases and Flowers (2013多校第二场线段树)
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题意摘自:http://blog.csdn.net/kdqzzxxcc/article/details/9474169 ORZZ
题意:给你N个花瓶,编号是0 到 N - 1 ,初始状态花瓶是空的,每个花瓶最多插一朵花。
然后有2个操作。操作1,a b c ,往在a位置后面(包括a)插b朵花,输出插入的首位置和末位置。
操作2,a b ,输出区间[a , b ]范围内的花的数量,然后全部清空。
很显然这是一道线段树。区间更新,区间求和,这些基本的操作线段树都可以logN的时间范围内完成。
操作2,很显然就是线段树的区间求和,求出[a , b]范围内的花朵的数量,区间更新,将整个区间全部变成0。
操作1,这里我们首先需要找出他的首位置和末位置,所以需要二分他的位置。
首先我们二分他的首位置, l = a , r = n ,在这个区间内二分,找出第一个0的位置,那就是该操作的首位置pos1。
然后再二分他的末位置,l = pos1 , r = n ,找到第c个0,就是该操作的末位置pos2,然后区间更新[pos1 ,pos2]全部置为1。
tmp = n - minn + 1 - query(minn,n,1) ; 表示【minn,n】区间内空花瓶的个数,如果需要插的花束多余tmp,则只能插tmp束花。
这题重点在于二分找位置,找了半天终于在别人的帮助下调试出来了............
#include <iostream>#include <algorithm>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <vector>#include <set>#include <queue>#include <stack>#include <climits>//形如INT_MAX一类的#define MAX 100050#define INF 0x7FFFFFFF#define REP(i,s,t) for(int i=(s);i<=(t);++i)#define ll long long#define mem(a,b) memset(a,b,sizeof(a))#define mp(a,b) make_pair(a,b)#define L(x) x<<1#define R(x) x<<1|1# define eps 1e-5//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂using namespace std;struct node { int r,l,mid; int lazy,sum;} edge[4*MAX];void up(int num) { edge[num].sum = edge[L(num)].sum + edge[R(num)].sum;}void build(int l,int r,int num) { edge[num].l = l; edge[num].r = r; edge[num].mid = (l+r) >> 1; edge[num].lazy = -1; if(l == r) { edge[num].sum = 0; return ; } build(l,edge[num].mid,L(num)); build(edge[num].mid+1,r,R(num)); // up(num);}void down(int num) { //if(edge[num].l == edge[num].r) return ; if(edge[num].lazy != -1) { edge[L(num)].lazy = edge[num].lazy; edge[R(num)].lazy = edge[num].lazy; edge[L(num)].sum = (edge[L(num)].r - edge[L(num)].l + 1) * edge[num].lazy; edge[R(num)].sum = (edge[R(num)].r - edge[R(num)].l + 1) * edge[num].lazy; edge[num].lazy = -1; }}void update(int l,int r,int num,int k) { if(l == edge[num].l && r == edge[num].r) { edge[num].lazy = k; edge[num].sum = (edge[num].r - edge[num].l + 1) * k; return ; } down(num); if(r <= edge[num].mid) { update(l,r,L(num),k); } else if(l > edge[num].mid) { update(l,r,R(num),k); } else { update(l,edge[num].mid,L(num),k); update(edge[num].mid + 1,r,R(num),k); } up(num);}int query(int l,int r,int num) { if(l == edge[num].l && r == edge[num].r) { return edge[num].sum; } down(num); if(r <= edge[num].mid) { return query(l,r,L(num)); } else if(l > edge[num].mid) { return query(l,r,R(num)); } else { return query(l,edge[num].mid,L(num)) + query(edge[num].mid + 1,r,R(num)); }}void test(int n) { for(int i=1; i<=3*n; i++) { printf("l:%d r:%d sum:%d lazy:%d\n",edge[i].l,edge[i].r,edge[i].sum,edge[i].lazy); }}int main() { int n,m,i,t; cin >> t; while(t --) { scanf("%d%d",&n,&m); int a,b,c; build(1,n,1); for(i=0; i<m; i++) { scanf("%d%d%d",&a,&b,&c); if(a == 1) { int low = b + 1,high = n,mid,minn = 2*n; if(n - low + 1 - query(low , n , 1) == 0) { printf("Can not put any one.\n"); continue; } while(low <= high) { mid = (low + high) >> 1; if(mid - (b + 1) + 1- query(b+1,mid,1) >= 1) { minn = min(minn,mid); high = mid - 1; } else { low = mid + 1; } } int tmp = n - minn + 1 - query(minn,n,1) ; if(c >= tmp) c = tmp; · low = minn,high = n; int maxx = INF; while(low <= high) { mid = (low + high) >> 1; tmp = mid - minn + 1 - query(minn,mid,1) ; if(tmp == c) { maxx = min(maxx,mid); high = mid - 1; } else if(tmp > c) { high = mid - 1; } else { low = mid + 1; } } printf("%d %d\n",minn-1,maxx-1); update(minn,maxx,1,1); } else { printf("%d\n",query(b+1,c+1,1)); update(b+1,c+1,1,0); } } puts(""); } return 0;}
这题重点在于二分找位置,找了半天终于在别人的帮助下调试出来了............
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