HDU1698 Just a Hook

Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15705    Accepted Submission(s): 7797

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


对应线段树的区间更新，代码有注释

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX=100010;
struct seg
{
    int l,r;
    int sum,mark;
}tree[MAX*4];
void pushdown(int k)    //向下传递，延迟操作
{
    int x=tree[k].r-tree[k].l+1;        //求出这个区间的长度
    tree[2*k].mark=tree[2*k+1].mark=tree[k].mark;   //把需要更新的值向下传递
    tree[2*k].sum=(x-(x>>1))*tree[k].mark;  //更新子节点的值
    tree[2*k+1].sum=(x>>1)*tree[k].mark;
    tree[k].mark=0;
}
void build(int l,int r,int k)
{
    tree[k].l=l;
    tree[k].r=r;
    tree[k].mark=0;
    if(l==r)
    {
        tree[k].sum=1;      //初始值是1
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,2*k);
    build(mid+1,r,2*k+1);
    tree[k].sum=tree[2*k].sum+tree[2*k+1].sum;
}
void insert(int l,int r,int k,int val)
{
    if(tree[k].l>=l&&tree[k].r<=r)     //修改的区间完全包含这个节点的区间
    {
        tree[k].mark=val;      //标记，如需要后面做延迟操作
        tree[k].sum=(tree[k].r-tree[k].l+1)*val;
        return;
    }
    if(tree[k].mark!=0)
        pushdown(k);    //延迟操作，把更新向下传递，只有在需要时才更新子节点，大大提高了效率
    int mid=(tree[k].l+tree[k].r)>>1;
    if(mid>=l)
        insert(l,r,2*k,val);
    if(mid<r)
        insert(l,r,2*k+1,val);
    tree[k].sum=tree[2*k].sum+tree[2*k+1].sum;
}
int query(int l,int r,int k)
{
    if(tree[k].r<l||tree[k].l>r)
        return 0;
    if(tree[k].l>=l&&tree[k].r<=r)
        return tree[k].sum;
    if(tree[k].mark!=0)
        pushdown(k);
    return query(l,r,2*k)+query(l,r,2*k+1);
}
int main()
{
    int t,n,m,x,y,flag=1,a;
    //freopen("in.txt","r",stdin);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        build(1,n,1);
        while(m--)
        {
            scanf("%d%d%d",&x,&y,&a);
            insert(x,y,1,a);
        }
        printf("Case %d: The total value of the hook is %d.\n",flag++,query(1,n,1));
    }
    return 0;
}