Hdu1698 just a hook (updated slot mark of Line Segment tree)

Just a hook Time Limit: 4000/2000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 22730 accepted submission (s): 11366


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 creating 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.

 
Source2008 "sunline Cup" national invitational contest
Recommendwangye | we have carefully selected several similar problems for you: 1542 1394 2795 1255
Statistic | submit | discuss | note Do not update the data directly to the bottom during the update process. It will time out. No need to try .. I have become a mouse ..

To use the lazy tag... is to update only to the interval at the time of update, and mark the lazy [] array. If the last updated vertex is used at the next update

Update the last lazy [] array marker

I want to lament .. ACM is too tired .. You can't do it now, and you cannot time out.

#include <stdio.h>struct node{int left,right,val,lazy,tag;}c[100000*3];void build_tree(int l,int r,int root){c[root].left=l;c[root].right=r;c[root].lazy=0;c[root].tag=0;if(l==r){c[root].val=1;return ;}int mid=(c[root].left+c[root].right)/2;build_tree(l,mid,root*2);build_tree(mid+1,r,root*2+1);c[root].val=c[root*2].val+c[root*2+1].val;}void update_tree(int l,int r,int x,int root){if(c[root].left==l&&r==c[root].right){c[root].val=(r-l+1)*x;c[root].lazy=1;c[root].tag=x;return ;}int mid=(c[root].left+c[root].right)/2;if(c[root].lazy==1){c[root].lazy=0;update_tree(c[root].left,mid,c[root].tag,root*2);update_tree(mid+1,c[root].right,c[root].tag,root*2+1);c[root].tag=0;}if(mid<l)update_tree(l,r,x,root*2+1);else if(mid>=r)update_tree(l,r,x,root*2);else{update_tree(l,mid,x,root*2);update_tree(mid+1,r,x,root*2+1);}c[root].val=c[root*2].val+c[root*2+1].val;}int main(){int ncase,n,k;scanf("%d",&ncase);for(int t=1;t<=ncase;t++){scanf("%d",&n);build_tree(1,n,1);scanf("%d",&k);while(k--){int a,b,x;scanf("%d %d %d",&a,&b,&x);update_tree(a,b,x,1);}printf("Case %d: The total value of the hook is %d.\n",t,c[1].val);}return 0;}

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Hdu1698 just a hook (updated slot mark of Line Segment tree)