hdu1698-Just a Hook (线段树区间更新)（延迟标记）

Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 32616    Accepted Submission(s): 15997

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.

 

题目大意：

有一条线段，每次可以把线段中的某个区间的价值改变，最后求这个线段的总价值。

解题思路：

这是一道关于线段树的区间更新题目，需要使用延迟标记，如果不使用会超时，和hdu1566题目很类似，可以在建树的时候不处理，然后我们在更新的时候标记并处理，能节约大量时间，具体可以参考代码。

ac代码：

#include <cstdio>  #include <algorithm>  using namespace std;  #define lson l , m , rt << 1  #define rson m + 1 , r , rt << 1 | 1  const int maxn = 111111;  int h , w , n;  int col[maxn<<2];  //一般数组开4倍 int sum[maxn<<2];  void PushUp(int rt) {  //递归求总价值     sum[rt] = sum[rt<<1] + sum[rt<<1|1];  }  void PushDown(int rt,int m) {  //处理延迟标记     if (col[rt]) {          col[rt<<1] = col[rt<<1|1] = col[rt];          sum[rt<<1] = (m - (m >> 1)) * col[rt];          sum[rt<<1|1] = (m >> 1) * col[rt];          col[rt] = 0;      }  }  void build(int l,int r,int rt) {  //建树     col[rt] = 0;      sum[rt] = 1;      if (l == r) return ;      int m = (l + r) >> 1;      build(lson);      build(rson);      PushUp(rt);  //回溯求总价值 }  void update(int L,int R,int c,int l,int r,int rt) {  //区间更新     if (L <= l && r <= R) {          col[rt] = c;          sum[rt] = c * (r - l + 1);          return ;      }      PushDown(rt , r - l + 1);  //处理标记     int m = (l + r) >> 1;      if (L <= m) update(L , R , c , lson);      if (R > m) update(L , R , c , rson);      PushUp(rt);  //回溯求总价值 }  int main() {      int T , n , m;      scanf("%d",&T);      for (int cas = 1 ; cas <= T ; cas ++) {          scanf("%d%d",&n,&m);          build(1 , n , 1);  //建树         while (m --) {              int a , b , c;              scanf("%d%d%d",&a,&b,&c);              update(a , b , c , 1 , n , 1);  //更新         }          printf("Case %d: The total value of the hook is %d.\n",cas , sum[1]);      }      return 0;  }  

题目网址：点击打开链接http://acm.hdu.edu.cn/showproblem.php?pid=1698

相似题目：点击打开链接http://blog.csdn.net/wang_heng199/article/details/75036010