【线段树五】HDU 1698 Just a Hook
时间：2014-04-29 10:32:46
收藏：0
阅读：457
Just a HookTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15093 Accepted Submission(s): 7489 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
1
10
2
1 5 2
5 9 3 Sample
Output
Case
1: The total value of the hook is 24. Source
Recommend
|
线段树应用---成段更新（lazy）
初始化：
操作1：
更新1--5 value=2
过程：
update(int from,int to,int value,int l,int r,int index)
1 5
2 1 10
1
Down(index,r-l+1);
sticks[0]=0;表示：上次没有更新你，你下面的也就不用更新，我往下找找吧
update(int from,int to,int value,int l,int r,int index)
1 5
2 1 mid=5 2
更新stick[2]=2, sum[2]=(r-l+1)*value=(5-1+1)*2=10
（这个【1,5】范围内先不用更新到底，假装更新完了）
return ;
操作2：
更新5--9 value=3
过程：
update(int from,int to,int value,int l,int r,int index)
5 9
3 1 10 1
Down(index,r-l+1);
sticks[0]=0;表示：上次没有更新你，你下面的也就不用更新，我往下找找吧
update(int from,int to,int value,int l,int r,int index)
5 9
3 1 mid=5 2
Down(index,r-l+1);
sticks[2]=2!=0
;表示，奥，上次更新过你了，你往下更新一下吧，我要使你孩子的值
-->sticks[4]=2，sticks[5]=2,更新和
sum[4]=6,sum[5]=4;
-->sticks[2]=0 重新置为0，你已经更新完了
等于没有更新你，更新了的是你孩子节点。
……
所以标记的作用在于不用每次更新到叶子节点，有需要的时候再往下更新。
#include <iostream>
#include<stdio.h>
using namespace std;
#define MAXN 100000
int
sticks[MAXN*4];
int
sum[MAXN*4];
void Up(int
index){
sum[index]=sum[index*2]+sum[index*2+1];
}
void Down(int
index,int m){
if(sticks[index]){
sticks[index*2]=sticks[index*2+1]=sticks[index];
sum[index*2]=(m-m/2)*sticks[index];
sum[index*2+1]=m/2*sticks[index];
sticks[index]=0;
}
}
void build(int
l,int r,int index){
sticks[index]=0;
sum[index]=1;
if(l==r){
return
;
}
int
mid=(r+l)/2;
build(l,mid,index*2);
build(mid+1,r,index*2+1);
Up(index);
}
void
update(int from,int to,int
value,int l,int r,int
index){
if(from<=l&&to>=r){
sticks[index]=value;
sum[index]=(r-l+1)*value;
return
;
}
Down(index,r-l+1);
int mid=(l+r)/2;
if(from<=mid)
update(from
,to,value,l,mid,index*2);
if(to>mid)
update(from,to,value,mid+1,r,index*2+1);
Up(index);
}
int
main()
{
int
T;
int N,Q;
scanf("%d",&T);
for(int i=1;i<=T;i++){
scanf("%d%d",&N,&Q);
build(1,N,1);
while(Q--){
int X,Y,Z;
scanf("%d%d%d",&X,&Y,&Z);
update(X,Y,Z,1,N,1);
}
printf("Case %d: The total value of the
hook is %d.\n",i , sum[1]);
}
return 0;
}
评论（0）