Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24474 Accepted Submission(s): 12194
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.
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
10
2
1 5 2
5 9 3
Sample Output
Case 1: The total value of the hook is 24.
Source
线段树 区间更新 区间求和
理解这个博客
要多刷
http://www.douban.com/note/273509745/
4.14重刷 更新模板 注意 pushdown的使用 有些题目 懒散标记不能堆积，得向下更新
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#define ll __int64
#define maxn 100000
using namespace std;
struct node
{
int l,r;
int add;
int sum;
}tree[*maxn];
int t;
int x,y,z;
int n,q;
int ans=;
void buildtree(int root,int left,int right)
{
tree[root].l=left;
tree[root].r=right;
tree[root].add=;
if(left==right)
{
tree[root].sum=;
return ;
}
int mid=(left+right)>>;
buildtree(root<<,left,mid);
buildtree(root<<|,mid+,right);
tree[root].sum=tree[root<<].sum+tree[root<<|].sum;
}
void pushdown(int root,int m)
{
tree[root<<].add=tree[root].add;
tree[root<<|].add=tree[root].add;
tree[root<<].sum=tree[root].add*(m-(m>>));
tree[root<<|].sum=tree[root].add*(m>>);
tree[root].add=;
}
void updata(int root,int left,int right,int c)
{
if(tree[root].l==left&&tree[root].r==right)
{
tree[root].add=c;
tree[root].sum=c*(right-left+);
return ;
}
if(tree[root].l==tree[root].r)
return ;
if(tree[root].add)
pushdown(root,tree[root].r-tree[root].l+);
int mid=(tree[root].l+tree[root].r)>>;
if(right<=mid)
updata(root<<,left,right,c);
else
{
if(left>mid)
updata(root<<|,left,right,c);
else
{
updata(root<<,left,mid,c);
updata(root<<|,mid+,right,c);
}
}
tree[root].sum=tree[root<<].sum+tree[root<<|].sum;
}
int query(int root,int left,int right)
{
if(tree[root].l==left&&tree[root].r==right)
{
return tree[root].sum;
}
if(tree[root].add)
pushdown(root,tree[root].r-tree[root].l+);
int mid=(tree[root].l+tree[root].r)>>;
if(right<=mid)
ans+=query(root<<,left,right);
else
{
if(left>mid)
ans+=query(root<<|,left,right);
else
{
ans+=query(root<<,left,mid);
ans+=query(root<<|,mid+,right);
}
}
return ans;
}
int main()
{
while(scanf("%d",&t)!=EOF)
{
for(int i=;i<=t;i++)
{
scanf("%d",&n);
buildtree(,,n);
scanf("%d",&q);
for(int j=;j<=q;j++)
{
scanf("%d %d %d",&x,&y,&z);
updata(,x,y,z);
}
ans=;
printf("Case %d: The total value of the hook is %d.\n",i,query(,,n));
}
} return ;
}
旧板子
#include<bits/stdc++.h>
using namespace std;
#define maxn 100005
struct ss
{
int l,r;
int sum;
int tag;
}tr[4*maxn];
void build (int k,int s,int t)
{
tr[k].l=s;
tr[k].r=t;
tr[k].tag=0;
if(s==t)
{
tr[k].sum=1;
return ;
}
int mid=(s+t)>>1;
build(k<<1,s,mid);
build(k<<1|1,mid+1,t);
tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum;
//cout<<tr[k].sum<<endl;
}
void pushdown(int k,int m)//向下更新节点
{
tr[k<<1].tag=tr[k].tag;
tr[k<<1|1].tag=tr[k].tag;
tr[k<<1].sum=tr[k].tag*(m-(m>>1));
tr[k<<1|1].sum=tr[k].tag*(m>>1);
tr[k].tag=0;
}
void update (int k,int s,int t,int exm)
{
if(tr[k].l==s&&tr[k].r==t)
{
tr[k].tag=exm;
tr[k].sum=exm*(t-s+1);
return ;
}
if(tr[k].l==tr[k].r)
return ;
if(tr[k].tag)
pushdown(k,tr[k].r-tr[k].l+1);
int mid=(tr[k].l+tr[k].r)>>1;
if(t<=mid)
update(k<<1,s,t,exm);
else if(s>mid)
update(k<<1|1,s,t,exm);
else
{
update(k<<1,s,mid,exm);
update(k<<1|1,mid+1,t,exm);
}
tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum;
}
int ask(int k,int s,int t)
{
if(tr[k].l==s&&tr[k].r==t)
{
return tr[k].sum;
}
if(tr[k].tag) pushdown(k,tr[k].r-tr[k].l+1);
int mid=(tr[k].l+tr[k].r)>>1;
int ans=0;
if(t<=mid)
ans+= ask(k<<1,s,t);
else
if(s>mid)
ans+=ask(k<<1|1,s,t);
else
{
ans+=ask(k<<1,s,mid);
ans+=ask(k<<1|1,mid+1,t);
}
return ans;
}
int t;
int n;
int m;
int qq,ww,ee;
//int a[maxn];
int main()
{
while(scanf("%d",&t)!=EOF)
{
for(int i=1;i<=t;i++)
{ scanf("%d",&n);
build(1,1,n);
scanf("%d",&m);
for(int j=1;j<=m;j++)
{
scanf("%d%d%d",&qq,&ww,&ee);
update(1,qq,ww,ee);
}
printf("Case %d: The total value of the hook is %d.\n",i,ask(1,1,n));
}
}
return 0;
}