hdu 1698 线段树 区间更新
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Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 36313 Accepted Submission(s):
17713
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.
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.
#include <bits/stdc++.h>
using namespace std;
const int maxn=100000+10;
struct node
{
int l,r,sum,lazy;
}t[maxn<<2];
void build (int l,int r,int i)
{
t[i].r=r;
t[i].l=l;
t[i].lazy=0;
if(l==r)
{
t[i].sum=1;
return ;
}
int mid;
mid=(l+r)>>1;
build(l,mid,i<<1);
build(mid+1,r,i<<1|1);
t[i].sum=(t[i<<1].sum+t[i<<1|1].sum);
}
void pushdown (int i)
{
t[i<<1].lazy = t[i].lazy;
t[i<<1|1].lazy= t[i].lazy;
int len=t[i].r-t[i].l+1;
t[i<<1].sum= (len-len/2)*t[i].lazy;
t[i<<1|1].sum=(len/2)*t[i].lazy;
t[i].lazy=0;
}
void update(int l,int r,int i,int z)
{
if(l<=t[i].l && r>=t[i].r)
{
t[i].lazy=z;
t[i].sum=(t[i].r-t[i].l+1)*z;
}
else
{
if(t[i].lazy) pushdown(i);
int mid=(t[i].r+t[i].l)/2;
if(l<=mid)
update(l,r,i<<1,z);
if(r>=mid+1)
update(l,r,i<<1|1,z);
t[i].sum=(t[i<<1].sum+t[i<<1|1].sum);
}
}
int main()
{
int t1,a,b,z,n,Q;
scanf("%d",&t1);
for(int i=1;i<=t1;i++)
{
scanf("%d",&n);
build(1,n,1);
scanf("%d",&Q);
for(int j=0;j<Q;j++)
{
scanf("%d%d%d",&a,&b,&z);
update(a,b,1,z);
//for(int k=1;k<=19;k++)
//if(t[k].l)
//printf("%d %d=%d lazy:%d\\n",t[k].l,t[k].r,t[k].sum,t[k].lazy);
}
printf("Case %d: The total value of the hook is %d.\\n",i,t[1].sum);
}
return 0;
}
// pushdown 和 update 函数bug:
修改后ac:
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