HDU 1698（线段树区间更新）

Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21444    Accepted Submission(s): 10735

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

 

好，，dota题，。。

题意： 给你一个1到n的区间，区间的初始值为一，然后下面q个操作，每个操作包含3个三个整数x,y,z，表示区间x,y里面的值修改为z。然后经过q个操作之后，整个区间的权值是多少。

显然是一道线段树区间更新的题目。、（需要用到lazy标记）

吐槽：    屠夫出强袭并不合理啊。。。

#include <cstdio>#include <algorithm>#include <iostream>using namespace std;#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1#define root 1 , N , 1#define LL long longconst int maxn = 111111;LL add[maxn<<2];LL sum[maxn<<2];void PushUp(int rt){    sum[rt] = sum[rt<<1] + sum[rt<<1|1];}void PushDown(int rt,int m){    if (add[rt])   //lazy思想。    {        add[rt<<1]=add[rt<<1|1]=add[rt];        sum[rt<<1]= add[rt] * (m - (m >> 1));        sum[rt<<1|1]= add[rt] * (m >> 1);        add[rt] = 0;    }}void build(int l,int r,int rt){    add[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)    {        add[rt]= c;        sum[rt]= (LL)c * (r - l + 1);        return ;    }    PushDown(rt , r - l + 1);    int m = (l + r) >> 1;    if (L <= m) update(L , R , c , lson);    if (m < R) update(L , R , c , rson);    PushUp(rt);}int main(){    int N,m;    int Q;    cin>>Q;    int dd=0;    while (Q--)    {        scanf("%d%d",&N,&m);        build(root);        int a,b,c;        while(m--)        {            scanf("%d%d%d",&a,&b,&c);            update(a,b,c,root);        }        cout<<"Case "<<++dd<<": The total value of the hook is ";        cout<<sum[1]<<"."<<endl;//整个区间和在节点1上。    }    return 0;}