Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14516 Accepted Submission(s): 7176
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.
思路：线段树，设计节点的时候不用记录该节点区间里面的sum，只需要记录每个区间里面的kind就行，（sum=R-L+1)*kind)，这样更新的时候就看当前节点区间和插入的区间[L,R]的关系：1，如果当前节点区间的kind和插入区间的kind相同就直接结束UpdateTree（）函数，不需要处理；2，否则，如果当前节点区间与插入区间正好匹配，把那么就直接把当前节点区间的kind置为插入区间的kind，结束UpdateTree（）函数；3，如果上述两条均不满足，则：（1）如果当前节点区间是纯色的（就是只有一种kind），那么该节点区间就变成不是纯色的了，（不止一种kind），那么就将该节点区间的kind变成0，同时注意更新其子区间的kind，因为我们并没有处理当前节点区间，而是把它转移到其子区间了。然后我们应该找该节点区间的子区间，直到满足1和2的条件（即是找到纯色区间或找到平匹配的区间）。（2）如果当前节点区间不是纯色那么就递归寻找其子区间，直到满足1,2条件。
1 /************************************************************************* 2 > File Name: Hook.cpp 3 > Author: wangzhili 4 > Mail: wangstdio.h@gmail.com 5 > Created Time: 2014/2/27 星期四 12:40:45 6 ************************************************************************/ 7 #include<iostream> 8 #include<cstdio> 9 using namespace std; 10 #define MAX 100000 11 class Tree_Node 12 { 13 public: 14 int left; 15 int right; 16 int mid; 17 int kind; 18 Tree_Node() 19 { 20 kind = 1; 21 } 22 }; 23 Tree_Node node[4*MAX]; 24 void BuildTree(int k, int l, int r) 25 { 26 node[k].left = l; 27 node[k].right = r; 28 node[k].kind = 1; 29 node[k].mid = (l + r) >> 1; 30 if(l == r) 31 return ; 32 int mid = (l + r) >> 1; 33 BuildTree(k << 1, l, mid); 34 BuildTree(k << 1|1, mid + 1, r); 35 } 36 37 void UpdateTree(int k, int l, int r, int kind) 38 { 39 if(node[k].left == l && node[k].right == r) 40 { 41 node[k].kind = kind; 42 return ; 43 } 44 if(node[k].kind == kind) 45 return ; 46 if(node[k].kind) 47 { 48 node[k << 1].kind = node[k].kind; 49 node[k << 1|1].kind = node[k].kind; 50 } 51 node[k].kind = 0; 52 if(node[k].mid < l) 53 UpdateTree(k << 1|1, l, r, kind); 54 else if(node[k].mid >= r) 55 UpdateTree(k << 1, l, r, kind); 56 else 57 { 58 UpdateTree(k << 1, l, node[k].mid, kind); 59 UpdateTree(k << 1|1, node[k].mid + 1, r,kind); 60 } 61 } 62 63 int GetSum(int k) 64 { 65 if(node[k].kind) 66 return (node[k].right-node[k].left + 1) * node[k].kind; 67 return GetSum(k << 1) + GetSum(k << 1|1); 68 } 69 70 int main(int argc, char const *argv[]) 71 { 72 int cnt, T, n, Q, i; 73 int l, r, kind; 74 cnt = 0; 75 // freopen("in.c", "r", stdin); 76 scanf("%d", &T); 77 while(T--) 78 { 79 cnt ++; 80 scanf("%d%d", &n, &Q); 81 BuildTree(1, 1, n); 82 for(i = 0; i < Q; i ++) 83 { 84 scanf("%d%d%d", &l, &r, &kind); 85 UpdateTree(1, l, r, kind); 86 } 87 printf("Case %d: The total value of the hook is %d. ",cnt, GetSum(1)); 88 } 89 return 0; 90 }