hdu 1698:Just a Hook（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): 15129 Accepted Submission(s): 7506
Total Submission(s): 15129 Accepted Submission(s): 7506
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.
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.
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
Recommend
线段树，区间更新。
与点更新不同的是，这是对某一区间的值进行更新。线段树每一个节点代表一个区间，节点中存储的值代表这个区间的金属成色（1-铜，2-银，3-金），如果这个区间还有其它成色，即杂色金属，则值为-1。
代码：
1 #include <stdio.h>
2 #define MAXSIZE 100000
3 struct Node{
4 int left,right;
5 int n;
6 };
7 Node a[MAXSIZE*3+1];
8 void Init(Node a[],int L,int R,int d) //初始化线段树
9 {
10 if(L==R){ //当前节点没有儿子节点，即递归到叶子节点。递归出口
11 a[d].left = L;
12 a[d].right = R;
13 a[d].n = 1;
14 return ;
15 }
16
17 int mid = (L+R)/2; //初始化当前节点
18 a[d].left = L;
19 a[d].right = R;
20 a[d].n = 1;
21
22 Init(a,L,mid,d*2); //递归初始化当前节点的儿子节点
23 Init(a,mid+1,R,d*2+1);
24
25 }
26 void Update(Node a[],int L,int R,int d,int x) //对区间[L,R]插入值x，从节点d开始更新。
27 {
28 if(a[d].n==x) //颜色相符，直接返回
29 return ;
30 if(L==a[d].left && R==a[d].right){ //插入的区间匹配，则直接修改该区间值
31 a[d].n = x;
32 return ;
33 }
34 if(a[d].n!=-1){ //是纯色
35 a[2*d].n=a[2*d+1].n=a[d].n;
36 a[d].n=-1;
37 }
38 int mid = (a[d].left + a[d].right)/2;
39 if(R<=mid){ //中点在右边界R的右边，则应该插入到左儿子
40 Update(a,L,R,d*2,x);
41 }
42 else if(mid<L){ //中点在左边界L的左边，则应该插入到右儿子
43 Update(a,L,R,d*2+1,x);
44 }
45 else { //否则，中点在待插入区间的中间
46 Update(a,L,mid,d*2,x);
47 Update(a,mid+1,R,d*2+1,x);
48 }
49 }
50 int Query(Node a[],int L,int R,int d) //查询区间[L,R]的值，从节点d开始查询
51 {
52 if(a[d].n==-1) //杂色
53 return Query(a,L,R,2*d) + Query(a,L,R,2*d+1);
54 else
55 return (a[d].right - a[d].left + 1)*a[d].n;
56 }
57 int main()
58 {
59 int Case,i,T,n,q,A,B,x;
60 scanf("%d",&T);
61
62 for(Case=1;Case<=T;Case++){
63 scanf("%d%d",&n,&q);
64
65 Init(a,1,n,1); //初始化
66
67 //for(i=1;i<=n;i++){ //输入
68 // Update(a,i,i,1,1);
69 //}
70
71 for(i=1;i<=q;i++){
72 scanf("%d%d%d",&A,&B,&x);
73 Update(a,A,B,1,x);
74 }
75 printf("Case %d: The total value of the hook is %d.\n",Case,Query(a,1,n,1));
76 }
77 return 0;
78 }
Freecode : www.cnblogs.com/yym2013
