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.
InputThe 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.
OutputFor 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. 解题思路：题目大意：在DOTA里有一个英雄帕琪（不玩，不知道叫啥，还是屠夫来着），他有一个肉钩，巨长，然后默认都是黄铜做的，每一节的价值都为1。然后他要改变钩子连续链节，换成别的材质。然后题目开始输入样例的个数，然后是钩子的链节数，接下来是操作的次数。后面就是更改了，1是黄铜，2是白银，3是金，价值也是这个数。
有一根棍，现在要给他表面镀上一层金属，（本身这根棍是铜的）
铜的价值是1
银得价值是2
金的价值是3
求最后的价值是多少。
这是区间值的修改，所以不同于之前的区间累加，lazy值是不用累加的；代码如下：
1 #include<iostream>
2 #include<stdio.h>
3 using namespace std;
4
5 int T;
6 int N;
7 int M;
8 const int MAXN = 1000005;
9 int a[MAXN];
10 int tree[MAXN*4];
11 int lazy[MAXN*4];
12 int x , y , k ;
13 int num = 0;
14 void push_down(int l ,int r ,int rt)
15 {
16 int m = (l+r)/2;
17
18 if(lazy[rt])
19 {
20 tree[rt*2] = lazy[rt]*(m-l+1);
21 tree[rt*2+1] = lazy[rt]*(r-m);
22 lazy[rt*2] = lazy[rt];
23 lazy[rt*2+1] = lazy[rt];
24 lazy[rt] = 0;
25 }
26 }
27 void bulid_tree(int l ,int r ,int rt)
28 {
29 lazy[rt] = 0;
30 if(l==r)
31 {
32 tree[rt] = a[l];
33 return ;
34 }
35 int m = (l+r)/2;
36 bulid_tree(l,m,rt*2);
37 bulid_tree(m+1,r,rt*2+1);
38 tree[rt] = tree[rt*2]+tree[rt*2+1];
39 }
40
41 void Update(int x ,int y ,int value, int l ,int r ,int rt)
42 {
43 if(x<=l&&r<=y)
44 {
45 tree[rt] = value*(r-l+1);
46 lazy[rt] = value;
47 return ;
48 }
49 push_down(l,r,rt);
50 int m = (l+r)/2;
51
52 if(x<=m)
53 {
54 Update(x,y,value,l,m,rt*2);
55 }
56
57 if(y>m)
58 {
59 Update(x,y,value,m+1,r,rt*2+1);
60 }
61 tree[rt] = tree[rt*2] +tree[rt*2+1];
62 }
63 int main()
64 {
65 scanf("%d",&T);
66 while(T--)
67 {
68 num++;
69 scanf("%d",&N);
70 for(int i = 1 ; i <= N ;i++)
71 {
72 a[i] = 1; //首先全部是铜，值为1；
73 }
74 bulid_tree(1,N,1);
75 scanf("%d",&M);
76 while(M--)
77 {
78 scanf("%d%d%d",&x,&y,&k);
79 Update(x,y,k,1,N,1);
80 }
81 printf("Case %d: The total value of the hook is %d.\n",num,tree[1]); //根实际上就是总和；
82 }
83 return 0;
84 }
原文地址：https://www.cnblogs.com/yewanting/p/10801342.html