题目链接:
http://poj.org/problem?id=3667
题意:
有一个线段,从1到n,下面m个操作,操作分两个类型,以1开头的是查询操作,以2开头的是更新操作
1 w 表示在总区间内查询一个长度为w的可用区间,并且要最靠左,能找到的话返回这个区间的左端点并占用了这个区间,找不到返回0
2 a len , 表示从单位a开始,清除一段长度为len的区间(将其变为可用,不被占用),不需要输出
题解:
线段树– 区间更新 区间合并 区间查询
lazy为-1表示没有更新到这个区间,所以不用pushdown; 为0表示这个区间没有被占用,所以传下去的长度就是当前区间的一半,左右各一半; 为1表示这个区间被占用了,给儿子区间的长度传成0就好了。
注意每次更新都要pushdown的是当前区间的长度
http://www.cnblogs.com/scau20110726/archive/2013/05/07/3065418.html 写的挺好
http://www.cnblogs.com/yewei/archive/2012/05/05/2484471.html 参考代码
代码:
1 #include <stdio.h> 2 #include <algorithm> 3 #include <cstring> 4 using namespace std; 5 typedef long long ll; 6 #define MS(a) memset(a,0,sizeof(a)) 7 #define MP make_pair 8 #define PB push_back 9 const int INF = 0x3f3f3f3f; 10 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL; 11 inline ll read(){ 12 ll x=0,f=1;char ch=getchar(); 13 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 14 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 15 return x*f; 16 } 17 ////////////////////////////////////////////////////////////////////////// 18 const int maxn = 1e5+10; 19 20 struct node{ 21 int l,r,lm,rm,mm,len,lazy; 22 void update(int val,int len){ 23 lazy = val; 24 if(val) lm=rm=mm = 0; 25 else lm=rm=mm = len; 26 } 27 }tree[maxn<<2]; 28 29 void pushup(int rt){ 30 tree[rt].lm = tree[rt<<1].lm; 31 tree[rt].rm = tree[rt<<1|1].rm; 32 tree[rt].mm = max(max(tree[rt<<1].mm,tree[rt<<1|1].mm),tree[rt<<1].rm+tree[rt<<1|1].lm); 33 if(tree[rt<<1].lm == tree[rt<<1].len) tree[rt].lm = tree[rt<<1].lm + tree[rt<<1|1].lm; 34 if(tree[rt<<1|1].rm == tree[rt<<1|1].len) tree[rt].rm = tree[rt<<1].rm+tree[rt<<1|1].rm; 35 } 36 37 void pushdown(int rt, int len){ 38 int lazyval = tree[rt].lazy; 39 if(lazyval != -1){ 40 tree[rt<<1].lazy = tree[rt<<1|1].lazy = tree[rt].lazy; 41 if(lazyval) { 42 tree[rt<<1].lm=tree[rt<<1].rm=tree[rt<<1].mm = 0; 43 tree[rt<<1|1].lm=tree[rt<<1|1].rm=tree[rt<<1|1].mm = 0; 44 }else{ 45 tree[rt<<1].lm = tree[rt<<1].rm = tree[rt<<1].mm = len-(len/2); 46 tree[rt<<1|1].lm = tree[rt<<1|1].rm = tree[rt<<1|1].mm = len/2; 47 } 48 tree[rt].lazy = -1; 49 } 50 } 51 52 void build(int rt,int l,int r){ 53 tree[rt].l = l, tree[rt].r = r; 54 tree[rt].lm = tree[rt].rm = tree[rt].mm = tree[rt].len = r-l+1; 55 tree[rt].lazy = -1; 56 if(l != r){ 57 int mid = (l+r)/2; 58 build(rt<<1,l,mid); 59 build(rt<<1|1,mid+1,r); 60 pushup(rt); 61 } 62 } 63 64 void update(int l,int r,int val,int rt){ 65 int L = tree[rt].l, R = tree[rt].r; 66 if(l<=L && R<=r){ 67 tree[rt].update(val,R-L+1); 68 return ; 69 } 70 pushdown(rt,R-L+1); 71 int mid = (L+R)/2; 72 if(l<=mid) update(l,r,val,rt<<1); 73 if(r>mid) update(l,r,val,rt<<1|1); 74 pushup(rt); 75 } 76 77 ll query(int rt,int len){ 78 int L = tree[rt].l, R = tree[rt].r; 79 if(L==R) 80 return L; 81 pushdown(rt,R-L+1); 82 if(tree[rt<<1].mm >= len) 83 return query(rt<<1,len); 84 else if(tree[rt<<1].rm+tree[rt<<1|1].lm >= len) 85 return tree[rt<<1].r-tree[rt<<1].rm+1; 86 else if(tree[rt<<1|1].mm >= len) 87 return query(rt<<1|1,len); 88 else 89 return 0; 90 } 91 92 int main(){ 93 int n = read(), m = read(); 94 build(1,1,n); 95 while(m--){ 96 int op = read(); 97 if(op == 1){ 98 int x = read(); 99 int ans = query(1,x); 100 printf("%d ",ans); 101 if(ans) 102 update(ans,ans+x-1,1,1); 103 }else{ 104 int x=read(),len=read(); 105 update(x,x+len-1,0,1); 106 } 107 } 108 109 return 0; 110 }