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
一般的线段树区间修改。但是这题有个坑处，就是区间值是直接替换而非增加值，所以pushdown的时候一定要把子节点的值也更新。←因为这个被坑了半小时青春
1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5 #include<cmath>
6 #define ls l,mid,rt<<1
7 #define rs mid+1,r,rt<<1|1
8 using namespace std;
9 const int mxn=200000;
10 int n,m;
11 int ans=0;
12 struct tree{
13 int sum;
14 int add;
15 }tr[mxn*4];
16 void Build(int l,int r,int rt){
17 if(l==r){
18 tr[rt].sum=1;
19 tr[rt].add=0;
20 return;
21 }
22 int mid=(l+r)>>1;
23 Build(ls);
24 Build(rs);
25 tr[rt].sum=tr[rt<<1].sum+tr[rt<<1|1].sum;
26 return;
27 }
28 void change(int L,int R,int x,int l,int r,int rt){
29 if(tr[rt].add){
30 int mid=(l+r)>>1;
31 tr[rt<<1].sum=(mid-l+1)*tr[rt].add;//更新下面的值很重要！
32 tr[rt<<1|1].sum=(r-mid)*tr[rt].add;//更新！
33 tr[rt<<1].add=tr[rt<<1|1].add=tr[rt].add;
34 tr[rt].add=0;
35 }
36 if(L<=l && r<=R){
37 tr[rt].sum=(r-l+1)*x;
38 tr[rt].add=x;
39 return;
40 }
41 int mid=(l+r)>>1;
42 if(L<=mid)change(L,R,x,ls);
43 if(R>mid)change(L,R,x,rs);
44 tr[rt].sum=tr[rt<<1].sum+tr[rt<<1|1].sum;
45 return;
46 }
47
48 int main(){
49 int T;
50 scanf("%d",&T);
51 for(int ro=1;ro<=T;ro++){
52 memset(tr,0,sizeof tr);
53 scanf("%d",&n);
54 Build(1,n,1);
55 scanf("%d",&m);
56 int i,j;
57 int x,y,z;
58 for(i=1;i<=m;i++){
59 scanf("%d%d%d",&x,&y,&z);
60 change(x,y,z,1,n,1);
61 }
62 printf("Case %d: The total value of the hook is %d.\n",ro,tr[1].sum);
63
64 }
65 return 0;
66 }