HDU(1698),线段树区间更新

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1698

区间更新重点在于懒惰标记。

当你更新的区间就是整个区间的时候，直接sum[rt] = c*(r-l+1);col[rt] = c；后面的子区间就不管了，当你下次更新某一个区间的时候，把col[rt]从顶往下推（也没有推到底），推到合适的位置，刚好这个位置是我要更新的区间的子区间(可能被横跨了)就停下来。

在这里感谢网上的大牛，感谢杰哥，彬哥。我才能AC。

Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27760    Accepted Submission(s): 13778

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

1 10 2 1 5 2 5 9 3

 

Sample Output

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

 

Source

2008 “Sunline Cup” National Invitational Contest 

#include <stdio.h>
#include <algorithm>

using namespace std;

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

const int maxn = 100010;

int sum[maxn<<2];
int col[maxn<<2];

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]; //往下推
        sum[rt<<1] = (m-(m>>1))*col[rt];    //子区间更新
        sum[rt<<1|1] = (m>>1)*col[rt];  
        col[rt] = 0;        //该点已经推了
    }
}

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()
{
    int T,n,m;
    scanf("%d",&T);
    for(int cases=1;cases<=T;cases++)
    {
        scanf("%d%d",&n,&m);
        build(1,n,1);
        while(m--)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            update(a,b,c,1,n,1);
        }
        printf("Case %d: The total value of the hook is %d.
",cases,sum[1]);
    }
    return 0;
}