HDU1698 Just a Hook 线段树入门题复习

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1698

Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27230    Accepted Submission(s): 13526

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.

思路：第一次写的时候每一次操作都去更新一个sum数组。TLE了。。

   所以要用个标记数组，记录最后一次更新的数值。

代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int maxn=100005;int sum[4*maxn];int col[4*maxn];//延迟数组void buildtree(int l,int r,int pos){  sum[pos]=0;  col[pos]=0;  if(l==r){sum[pos]=1;return;}  else  {    int mid=(l+r)>>1;    buildtree(l,mid,pos<<1);    buildtree(mid+1,r,pos<<1|1);  }  sum[pos]=sum[pos<<1]+sum[pos<<1|1];}void update(int l,int r,int resl,int resr,int pos,int value){  if(l==r){sum[pos]=value;return;}  if(resl<=l&&r<=resr)//属于这个区间的话  直接打延迟标记   {    col[pos]=value;    sum[pos]=(r-l+1)*value;    //col[pos<<1]=col[pos<<1|1]=col[pos];//延迟标记送给儿子们    return;  }  //延迟标记送给儿子们了  int mid=(l+r)>>1;  if(col[pos])  {    col[pos<<1]=col[pos<<1|1]=col[pos];    sum[pos<<1]=col[pos]*(mid-l+1);    sum[pos<<1|1]=col[pos]*(r-mid);    col[pos]=0;//儿子已经更新完毕   }  if(resl<=mid)update(l,mid,resl,resr,pos<<1,value);  if(resr>mid)update(mid+1,r,resl,resr,pos<<1|1,value);  sum[pos]=sum[pos<<1]+sum[pos<<1|1];}int main(){  int T,n,m,a,b,c;  scanf("%d",&T);  for(int t=1;t<=T;t++)  {    scanf("%d",&n);    buildtree(1,n,1);    scanf("%d",&m);    while(m--)    {      scanf("%d%d%d",&a,&b,&c);      update(1,n,a,b,1,c);    }    //show(1,n,1);    printf("Case %d: The total value of the hook is %d.\n",t,sum[1]);  }}