HDU---1689
2016-07-23 15:34
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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.
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn = 100009;
int n,sum;
struct node
{
int l,r,n;
} a[maxn<<2];
void init(int l,int r,int i)
{
a[i].l = l;
a[i].r = r;
a[i].n = 1;
if(l!=r)
{
int mid = (l+r)/2;
init(l,mid,2*i);
init(mid+1,r,2*i+1);
}
}
void insert(int i,int x,int y,int m)
{
if(a[i].n == m)//相同则不用修改了
return ;
if(a[i].l == x && a[i].r == y)//找到了区间,直接更新
{
a[i].n = m;
return ;
}
if(a[i].n != -1)//如果该区间只有一种颜色
{
a[2*i].n = a[2*i+1].n = a[i].n;//由于后面必定对子树操作,所以更新子树的值等于父亲的值
a[i].n = -1;//由于该区域颜色与修改不同,而且不是给定区域,所以该区域必定为杂色
}
//父区间为杂色时对所有子节点进行操作
int mid = (a[i].l+a[i].r)/2;
if(x>mid)
insert(2*i+1,x,y,m);
else if(y<=mid)
insert(2*i,x,y,m);
else
{
insert(2*i,x,mid,m);
insert(2*i+1,mid+1,y,m);
}
}
int find(int i)//区间求和
{
if(a[i].n != -1)//纯色直接算这个区间
return (a[i].r - a[i].l+1)*a[i].n;
else//不存则左右子树去找
return find(i*2)+find(i*2+1);
}
int main()
{
int T,k,x,y,m;
int t = 1;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
init(1,n,1);
while(k--)
{
scanf("%d%d%d",&x,&y,&m);
insert(1,x,y,m);
}
printf("Case %d: The total value of the hook is %d.\n",t++,find(1));
}
return 0;
}
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.
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn = 100009;
int n,sum;
struct node
{
int l,r,n;
} a[maxn<<2];
void init(int l,int r,int i)
{
a[i].l = l;
a[i].r = r;
a[i].n = 1;
if(l!=r)
{
int mid = (l+r)/2;
init(l,mid,2*i);
init(mid+1,r,2*i+1);
}
}
void insert(int i,int x,int y,int m)
{
if(a[i].n == m)//相同则不用修改了
return ;
if(a[i].l == x && a[i].r == y)//找到了区间,直接更新
{
a[i].n = m;
return ;
}
if(a[i].n != -1)//如果该区间只有一种颜色
{
a[2*i].n = a[2*i+1].n = a[i].n;//由于后面必定对子树操作,所以更新子树的值等于父亲的值
a[i].n = -1;//由于该区域颜色与修改不同,而且不是给定区域,所以该区域必定为杂色
}
//父区间为杂色时对所有子节点进行操作
int mid = (a[i].l+a[i].r)/2;
if(x>mid)
insert(2*i+1,x,y,m);
else if(y<=mid)
insert(2*i,x,y,m);
else
{
insert(2*i,x,mid,m);
insert(2*i+1,mid+1,y,m);
}
}
int find(int i)//区间求和
{
if(a[i].n != -1)//纯色直接算这个区间
return (a[i].r - a[i].l+1)*a[i].n;
else//不存则左右子树去找
return find(i*2)+find(i*2+1);
}
int main()
{
int T,k,x,y,m;
int t = 1;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
init(1,n,1);
while(k--)
{
scanf("%d%d%d",&x,&y,&m);
insert(1,x,y,m);
}
printf("Case %d: The total value of the hook is %d.\n",t++,find(1));
}
return 0;
}
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