hdu1698 Just a Hook 解題報告

Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11844    Accepted Submission(s): 5882


Problem DescriptionIn 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

11021 5 25 9 3

 

Sample Output

Case 1: The total value of the hook is 24.這題用到了線段樹的延遲標記，其實，說簡單一點，就是這次查詢的時候，我先不變，等下次再查詢的時候，我就順便就把這些延遲的更新，就行了，這樣的話，能充分用到UPDAT（）這個函數，延遲標記，就是線段樹的精華，必須要理解！
#include <iostream>#include <stdio.h>using namespace std;#define N 500000int l[N<<2],flag[N<<2];//l就 是總合 flag標記狀態void build(int num ,int s,int e){      flag[num]=0;      l[num]=1;      if(s==e)            return ;      int mid=(s+e)>>1;      build(num<<1,s,mid);      build(num<<1|1,mid+1,e);      l[num]=l[num<<1]+l[num<<1|1];}void update(int num ,int s,int e,int a,int b,int c){      if(a<=s&&b>=e)      {            flag[num]=c;            l[num]=c*(e-s+1);            return ;      }      if(flag[num])//是0的話，不變，到下次來更新      {            flag[num<<1]=flag[num<<1|1]=flag[num];            l[num<<1]=(e-s+1-(e-s+1)/2)*flag[num];            l[num<<1|1]=((e-s+1)/2)*flag[num];            flag[num]=0;//查詢進去後一定要標記成不更新狀態      }      int mid=(s+e)/2;      if(mid>=a)            update(num<<1,s,mid,a,b,c);      if(mid<b)            update(num<<1|1,mid+1,e,a,b,c);      l[num]=l[num<<1]+l[num<<1|1];}int main(){    int t,tt,m,n,a,b,c;    scanf("%d",&t);    for(tt=1;tt<=t;tt++)    {          scanf("%d",&n);          build(1,1,n);          scanf("%d",&m);          while(m--)          {             scanf("%d%d%d",&a,&b,&c);             update(1,1,n,a,b,c);          }          printf("Case %d: The total value of the hook is %d.\n",tt,l[1]);    }    return 0;}