Just a Hook（线段树成段更新）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1698

Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19079    Accepted Submission(s): 9575

Problem Description

In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length.



Now Pudge wants to do some operations on the hook.

Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.
The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:

For each cupreous stick, the value is 1.
For each silver stick, the value is 2.
For each golden stick, the value is 3.

Pudge wants to know the total value of the hook after performing the operations.
You may consider the original hook is made up of cupreous sticks.

 

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases.
For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations.
Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.

 

Output

For each case, print a number in a line representing the total value of the hook after the operations. Use the format in the example.

 

Sample Input

11021 5 25 9 3

 

Sample Output

Case 1: The total value of the hook is 24.

 

Source

2008 “Sunline Cup” National Invitational Contest

 

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成段更新(通常这对初学者来说是一道坎),需要用到延迟标记(或者说懒惰标记),简单来说就是每次更新的时候不要更新到底,用延迟标记使得更新延迟到下次需要更新or询问到的时候。

My  AC code：

#include<iostream>#include<algorithm>#include<cmath>#include<cstring>#include<cstdio>#include<queue>#define MAXN 100010#define LL long longusing namespace std;struct node{int l,r;int tag;//存储懒惰记录的数值 int lazy;//记录是否有懒惰记录的标记 LL s;}tree[MAXN*4];void build(int rt,int s,int e){tree[rt].l=s;tree[rt].r=e;tree[rt].lazy=0;//初始时无延迟标记，故tag和lazy都为0 tree[rt].tag=0;if(s==e){tree[rt].s=1;return;//记得return }int mid=(s+e)/2;build(rt*2,s,mid);build(rt*2+1,mid+1,e);tree[rt].s=tree[rt*2].s+tree[rt*2+1].s;//回溯 }void update(int rt,int ul,int ur,int val){if(tree[rt].l==ul&&tree[rt].r==ur){tree[rt].lazy=1;//延迟标记 tree[rt].tag=val;//存储懒惰记录的数值 tree[rt].s=(ur-ul+1)*val;return;//记得return }int mid=(tree[rt].l+tree[rt].r)/2;if(tree[rt].lazy==1)//有延迟标记,要先往下压,更新之前因为"偷懒"没更新的节点 {tree[rt].lazy=0;//此节点在"敦促"不敢偷懒，故延迟标记取消 update(rt*2,tree[rt].l,mid,tree[rt].tag);update(rt*2+1,mid+1,tree[rt].r,tree[rt].tag);tree[rt].tag=0;//懒惰记录的数值传给下面子区间的节点了，故取消懒惰记录的数值 }if(mid>=ur) update(rt*2,ul,ur,val);else if(mid<ul) update(rt*2+1,ul,ur,val);else{update(rt*2,ul,mid,val);update(rt*2+1,mid+1,ur,val);}tree[rt].s=tree[rt*2].s+tree[rt*2+1].s;}int main(){int t,x,y,z,cas,n,q;scanf("%d",&t);cas=0;while(t--){cas++;scanf("%d",&n);build(1,1,n);scanf("%d",&q);while(q--){scanf("%d%d%d",&x,&y,&z);update(1,x,y,z);}printf("Case %d: The total value of the hook is %lld.\n",cas,tree[1].s);}return 0; }