描述
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.
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
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
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.
Case 1: The total value of the hook is 24.
题意
1-n的编号，初始为全价值都为1，有Q个询问，每次询问让你把[X,Y]变成价值Z
题解
线段树区间更新延迟标记，区间查询
代码
#include<stdio.h>
#include<string.h> const int N=1e5+; int sum[N<<],lazy[N<<],ans; void PushDown(int rt)
{
if(lazy[rt])
{
lazy[rt<<]=lazy[rt<<|]=lazy[rt];
lazy[rt]=;
}
}
void Update(int L,int R,int C,int l,int r,int rt)
{
if(L<=l&&r<=R)
{
lazy[rt]=C;
return;
}
int mid=(l+r)>>;
PushDown(rt);
if(L<=mid)Update(L,R,C,l,mid,rt<<);
if(R>mid)Update(L,R,C,mid+,r,rt<<|);
}
void Query(int l,int r,int rt)
{
if(lazy[rt]==-||lazy[rt]==)
{
ans+=r-l+;
return;
}
else if(lazy[rt]==)
{
ans+=*(r-l+);
return;
}
else if(lazy[rt]==)
{
ans+=*(r-l+);
return;
}
int mid=(l+r)>>;
if(l<=mid)Query(l,mid,rt<<);
if(r>mid)Query(mid+,r,rt<<|);
}
int main()
{
int t,n,q,x,y,z,o=;
scanf("%d",&t);
while(t--)
{
memset(lazy,-,sizeof lazy);
scanf("%d%d",&n,&q);
for(int i=;i<q;i++)
{
scanf("%d%d%d",&x,&y,&z);
Update(x,y,z,,n,);
}
ans=;
Query(,n,);
printf("Case %d: The total value of the hook is %d.\n",o++,ans);
}
return ;
}