hdoj 1698 Just a Hook

原题：
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
1
10
2
1 5 2
5 9 3

Sample Output
Case 1: The total value of the hook is 24.
中文：
憎恶可以丢出去一条锁链，锁链有n个节。锁链上的节可以有三种分别是铜、银、金，代表的值分别为1、2、3。初始时这个锁链全是铜的，现在给你m个区间替换操作，把一段的长度替换成另外一种材质的锁链。最后问你所有锁链的值的和是多少？

#include <bits/stdc++.h>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const int maxn=100001;int col[maxn<<2],sum[maxn<<2];//替换标记，总和void PushUp(int rt){    sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void PushDown(int rt,int m){    if(col[rt])    {        col[rt<<1]=col[rt<<1|1]=col[rt];        sum[rt<<1]=(m-(m>>1))*col[rt];        sum[rt<<1|1]=(m>>1)*col[rt];        col[rt]=0;    }}void build(int l,int r,int rt){    col[rt]=0;    sum[rt]=1;    if(l==r) return;    int m=(l+r)>>1;    build(lson);    build(rson);    PushUp(rt);}void update(int L,int R,int c,int l,int r,int rt){    if(L<=l&&r<=R)    {        col[rt]=c;        sum[rt]=c*(r-l+1);        return;    }    PushDown(rt,r-l+1);    int m=(l+r)>>1;    if(L<=m)        update(L,R,c,lson);    if(R>m)        update(L,R,c,rson);    PushUp(rt);}int main(){    ios::sync_with_stdio(false);    int T,n,m;    cin>>T;    for(int cas=1;cas<=T;cas++)    {        cin>>n>>m;        build(1,n,1);        while(m--)        {            int a,b,c;            cin>>a>>b>>c;            update(a,b,c,1,n,1);        }        cout<<"Case "<<cas<<": The total value of the hook is "<<sum[1]<<"."<<endl;    }    return 0;}

答案：
第一道线段树的区间查询问题，代码完全参考网上力荐的HH的模板，最初是看刘汝佳的那本白书，结果发现不是很好理解=_=。
个人理解线段树区间更改的所用到的延迟标记方法大概是这么一回事。如果每次修改或者操作一个区间[l,r]，那么线段树的子区间和也需要去修改，而且修改的是一个区间部分，几乎会涉及到的整个线段树的修改，时间复杂度很高。这里采用的延迟标记方法的大概思想是，每次在当前大区间上做更改，暂时先不去动子区间，若有查询操作或者是另外一个更改操作涉及到当前区间的时候，把上一次没有修改的子区间，顺道修改掉，这也是查询操作query(此题没有query）当中也会有向子区间传递更改操作的原因。
讲解可以参考此篇博客，非常棒。
http://blog.sina.com.cn/s/blog_a2dce6b30101l8bi.html