hdu1698 线段树(区间更新~将区间[x,y]的值替换为z)
Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 45600 Accepted Submission(s): 21730
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 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.
each case, print a number in a line representing the total value of the
hook after the operations. Use the format in the example.
10
2
1 5 2
5 9 3
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<cmath>
using namespace std;
const int maxn = + ;
int t, n, x, y, z, q, tre[maxn * ], laz[maxn * ], kase = ;
void push_up(int num)
{
tre[num] = tre[num * ] + tre[num * + ];
}
void pushdown(int num, int x, int y)
{
if (laz[num])
{
int mid = (x + y) / ;
laz[num * ] = laz[num * + ] = laz[num];
tre[num * ] = (mid - x + ) * laz[num];
tre[num * + ] = (y - mid) * laz[num];
laz[num] = ;
}
}
void build(int x, int y, int num)
{
int mid = (x + y) / ;
laz[num] = ;
if (x == y)
{
tre[num] = ; return;
}
build(x, mid, num * );
build(mid + , y, num * + );
push_up(num);
}
void update(int le, int ri, int z, int x, int y, int num)
{
int mid = (x + y) / ;
if (le <= x && y <= ri)
{
laz[num] = z;
tre[num] = z * (y - x + );//先更新节点,但不继续往下更新,节点的值恰好是节点所在子树的和
return;
}
pushdown(num, x, y);
if (le <= mid)
update(le, ri, z, x, mid, num * );
if (mid < ri)
update(le, ri, z, mid + , y, num * + );
push_up(num);
}
int query(int le, int ri, int x,int y,int num)
{
if (le == ri)
{
return tre[num];
}
pushdown(num, x, y);
int mid = (le + ri) / ;
if (x <= mid)
return query(le, mid, x, y, num * );
else
return query(mid + , ri, x, y, num * + );
}
int main()
{
scanf("%d", &t);
while (t--)
{
scanf("%d%d", &n, &q);
build(, n, );
while (q--)
{
scanf("%d%d%d", &x, &y, &z);//将[x,y]的值更新为z
update(x, y, z, , n, );
}
printf("Case %d: The total value of the hook is %d.\n", ++kase, tre[]);
}
return ;
}
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