HDU 3308 LCSI 【线段树】 与 poj 3667 hotel

听说这个题目是那套题目里最简单的线段树了。

序列a 4 3 9 7 6 8

下标i 1 2 3 4 5 6

区间结点要记录的三个信息：当前最长递增的序列值，从左端开始的最长序列值，到右端结束的最长序列值。

 

 

线段树

 

 】

 

代码：

//query 语句最值得细看----多段联合  这个递归从最后开始往前联合。一般都要从后往前比较好。因为最小单位可以确定其值了，当然这里不需要这么考虑。

void query(int p,int l,int r,node &set) { if(tree[p].lt==l&&tree[p].rt==r) { set = tree[p]; return ; } int m = (tree[p].lt+tree[p].rt)>>1; if(m>=r){query(p*2,l,r,set);} else if(l>m){query(p*2+1,l,r,set);} else { node t1,t2; query(p*2,l,m,t1); query(p*2+1,m+1,r,t2); set.m = max(t1.m,t2.m); set.lt = t1.lt; set.rt= t2.rt; set.l = t1.l; set.r = t2.r; if(a[t1.rt]<a[t2.lt]) { set.m = max(set.m,(t1.r+t2.l)); if(t1.l==(t1.rt-t1.lt+1)) { set.l = t1.l+t2.l; } if(t2.r ==(t2.rt-t2.lt+1)) { set.r = t1.r+t2.r; } } } }

 

完整代码：

//写的比较丑~~~#include<stdio.h>#include<stdlib.h>#include<string.h>#define size 100010#define max(x,y) (x)>(y)?(x):(y)typedef struct tree{int lt,rt,l,r;int m;}node;node tree[size*3+1];int a[size];void build(int p,int l,int r)//建立线段树{tree[p].lt=l;tree[p].rt=r;if(l==r){tree[p].l=1;tree[p].r=1;tree[p].m=1;return;}int m = (l+r)>>1;build(p*2,l,m);build(p*2+1,m+1,r);//printf("%(1) %d %d %d %d/n",tree[p*2+1].lt,tree[p*2+1].rt,tree[p*2].m,tree[p*2+1].m);if(a[tree[p*2].rt]<a[tree[p*2+1].lt]){tree[p].m = max((tree[p*2].r+tree[p*2+1].l) , max(tree[p*2].m,tree[p*2+1].m));if(tree[p*2+1].r==(tree[p*2+1].rt-tree[p*2+1].lt+1))tree[p].r=tree[p*2].r+tree[p*2+1].r;elsetree[p].r=tree[p*2+1].r;if(tree[p*2].l==(tree[p*2].rt-tree[p*2].lt+1))tree[p].l=tree[p*2].l+tree[p*2+1].l;elsetree[p].l=tree[p*2].l;}else{tree[p].m = max(tree[p*2].m,tree[p*2+1].m);tree[p].l=tree[p*2].l;tree[p].r=tree[p*2+1].r;}}//queryvoid query(int p,int l,int r,node &set){if(tree[p].lt==l&&tree[p].rt==r){set = tree[p];return ;}int m = (tree[p].lt+tree[p].rt)>>1;if(m>=r){query(p*2,l,r,set);}else if(l>m){query(p*2+1,l,r,set);}else{node t1,t2;query(p*2,l,m,t1);query(p*2+1,m+1,r,t2);set.m = max(t1.m,t2.m);set.lt = t1.lt;set.rt= t2.rt;set.l = t1.l;set.r = t2.r;if(a[t1.rt]<a[t2.lt]){set.m = max(set.m,(t1.r+t2.l));if(t1.l==(t1.rt-t1.lt+1)){set.l = t1.l+t2.l;}if(t2.r ==(t2.rt-t2.lt+1)){set.r = t1.r+t2.r;}}}}void updata(int p,int l,int r,int ta,int tc){if(l==r&&l==ta&&r==ta){a[ta]=tc;return ;}int m = (l+r)>>1;if(m>=ta){updata(p*2,l,m,ta,tc);}else{updata(p*2+1,m+1,r,ta,tc);}if(a[tree[p*2].rt]<a[tree[p*2+1].lt]){tree[p].m = max((tree[p*2].r+tree[p*2+1].l) , max(tree[p*2].m,tree[p*2+1].m));if(tree[p*2+1].r==(tree[p*2+1].rt-tree[p*2+1].lt+1))tree[p].r=tree[p*2].r+tree[p*2+1].r;elsetree[p].r=tree[p*2+1].r;if(tree[p*2].l==(tree[p*2].rt-tree[p*2].lt+1))tree[p].l=tree[p*2].l+tree[p*2+1].l;elsetree[p].l=tree[p*2].l;}else{tree[p].m = max(tree[p*2].m,tree[p*2+1].m);tree[p].l=tree[p*2].l;tree[p].r=tree[p*2+1].r;}}int main(){int t,i,n,m,x,y;char c[2];while(scanf("%d",&t)!=EOF){while(t--){memset(tree,0,sizeof(tree));scanf("%d%d",&n,&m);for(i=1;i<=n;i++)scanf("%d",&a[i]);build(1,1,n);//printf("%d",tree[1].m);//for(i=1;i<=6*n-1;i++)//printf("%d %d max = %d/n",tree[1].lt,tree[1].rt,tree[1].m);while(m--){scanf("%s",c);if(c[0]=='Q'){scanf("%d%d",&x,&y);y++;x++;node set;query(1,x,y,set);printf("%d/n",set.m);}else{scanf("%d%d",&x,&y);x++;updata(1,1,n,x,y);}}}}return 0;}/*110 107 7 3 4 5 9 10 8 1 8 res.lt=l.lt; res.rt=r.rt; res.mx=max(l.mx, r.mx); if(dt[im-1]<dt[im]) { int len=l.rt+r.lt; if(l.rt==im-il) res.lt=len; if(r.lt==ir-im) res.rt=len; if(len>res.mx) res.mx=len; }*/

 

来看看poj的hotel吧  比较相似。把思想搞懂了就好做了。

看下图希望能够更好理解。

两天的对题目和代码的细细分析，终于搞懂理解了，o(∩_∩)o 哈哈 

主要关键是往回来的时候当前结点下面的信息要被当前结点影响。当前结点区间全部被占下层的孩子就是更新。

记住找区间的时候

if(tree[p].l==l&&tree[p].r==r)

否则会去递归到底层，这样就时间并没有优化，线段树也不起作用了。

看代码：

 #include<iostream>using namespace std;#define max(x,y) ((x)>(y)?(x):(y))#define size 50010struct seg {int l,r,llen,rlen,m;};seg tree[size*4];int n;void down(int p){if(tree[p].m==(tree[p].r-tree[p].l+1)){tree[p*2].m=tree[p*2].llen=tree[p*2].rlen=(tree[p*2].r-tree[p*2].l+1);tree[p*2+1].m=tree[p*2+1].llen=tree[p*2+1].rlen=(tree[p*2+1].r-tree[p*2+1].l+1);}else if(tree[p].m==0){tree[p*2].m=tree[p*2].llen=tree[p*2].rlen=0;tree[p*2+1].m=tree[p*2+1].llen=tree[p*2+1].rlen=0;}}void spread(int p) {int mx = max(tree[p*2].m,tree[p*2+1].m);if((tree[p*2].r-tree[p*2].l+1)==tree[p*2].llen){tree[p].llen = tree[p*2].llen + tree[p*2+1].llen;}else{tree[p].llen = tree[p*2].llen;}if((tree[p*2+1].r-tree[p*2+1].l+1)==tree[p*2+1].rlen){tree[p].rlen = tree[p*2].rlen + tree[p*2+1].rlen;}else{tree[p].rlen = tree[p*2+1].rlen;}tree[p].m = max(mx,(tree[p*2].rlen+tree[p*2+1].llen));}//checkvoid build(int p,int l,int r){tree[p].l=l;tree[p].r=r;tree[p].llen=tree[p].rlen=tree[p].m=tree[p].r-tree[p].l+1;if(l==r){tree[p].m=1;tree[p].llen=1;tree[p].rlen=1;return ;}int m = (l+r)>>1;build(p*2,l,m);build(p*2+1,m+1,r);}void ins(int p,int l,int r){if(tree[p].l==l&&tree[p].r==r){tree[p].llen=tree[p].rlen=tree[p].m=0;return ;}down(p);int mid = (tree[p].l+tree[p].r)>>1;if(mid>=r){ins(p*2,l,r);}else if(mid<l){ins(p*2+1,l,r);}else{ins(p*2,l,mid);ins(p*2+1,mid+1,r);}spread(p);}void del(int p,int l,int r){if(tree[p].l==l&&tree[p].r==r){tree[p].m=tree[p].llen=tree[p].rlen=tree[p].r-tree[p].l+1;return ;}down(p);int mid = (tree[p].l+tree[p].r)>>1;if(mid>=r){del(p*2,l,r);}else if(mid<l){del(p*2+1,l,r);}else{del(p*2,l,mid);del(p*2+1,mid+1,r);}spread(p);}int search(int p,int ps){down(p);if(tree[p].m<ps)return -1;if(tree[p].m>=ps){if(tree[p*2].m>=ps){return search(p*2,ps);}else if((tree[p*2].rlen+tree[p*2+1].llen)>=ps){return tree[p*2].r-tree[p*2].rlen+1;}}return search(p*2+1,ps);}int main(){int m,t,x,y;while(scanf("%d%d",&n,&m)!=EOF){build(1,1,n);while(m--){scanf("%d",&t);if(t==1){scanf("%d",&x);int l = search(1,x);if(l!=-1){ins(1,l,l+x-1);printf("%d/n",l);}else{printf("0/n");}}else{scanf("%d%d",&x,&y);if((x+y-1)>n){del(1,x,n);}else{del(1,x,x+y-1);}}}}return 0;}