HDU 1698 Just a Hook（线段树的区间更新）

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.

先给出点的个数
后来给出要更新的区间的个数。后来给出要更新的区间和要把次区间变为的值。
在这里面较难的问题为区间的更新，要判断某的区间是不是杂色的问题。

下面是AC代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

struct node
{
    int left,right,val;
} c[300005];

void build_tree(int l,int r,int root)
{
    c[root].left=l;
    c[root].right=r;
    c[root].val=1;
    if(l!=r)
    {
        int mid=(c[root].left+c[root].right)/2;
        build_tree(l,mid,root*2);
        build_tree(mid+1,r,root*2+1);
    }
}

int search_tree(int root)
{
    if(c[root].val!=-1)//如果不杂色，那么这个区间内每个点的值都是一样的
    {
        return (c[root].right-c[root].left+1)*c[root].val;
    }
    else return search_tree(root*2)+search_tree(root*2+1);//如果杂色的话，区间内每个点的值不一定一样，所以要从左右儿子求起
}

void update_tree(int l,int r,int root,int x)
{
    if(c[root].val==x)
    {
        return;
    }
    if(c[root].left==l&&c[root].right==r)
    {
        c[root].val=x;
        return ;
    }
    if(c[root].val!=-1)
    {
        c[root*2].val=c[root*2+1].val=c[root].val;
        c[root].val=-1;//如果不杂色的话，标记为杂色
    }
    int mid=(c[root].left+c[root].right)/2;
    if(mid<l)
    {
        update_tree(l,r,root*2+1,x);
    }
    else if(mid>=r)
    {
        update_tree(l,r,root*2,x);
    }
    else
    {
        update_tree(l,mid,root*2,x);
        update_tree(mid+1,r,root*2+1,x);
    }
}

int main()
{
    int t,iCase=0,n,m,a,b,cost;
    scanf("%d",&t);
    while(t--)
    {
        iCase++;
        scanf("%d",&n);
        build_tree(1,n,1);
        scanf("%d",&m);
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d",&a,&b,&cost);
            update_tree(a,b,1,cost);
        }
        printf("Case %d: The total value of the hook is %d.\n",iCase,search_tree(1));
    }
    return 0;
}