Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15645 Accepted Submission(s): 7750
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.
1 10 2 1 5 2 5 9 3
Case 1: The total value of the hook is 24.
Invitational Contest
传送门：HDU 1698 Just a Hook
题目大意：给你一个区间[1...n]，每次替换[L,R]的值为c，问最后整个区间的和是多少。
题目分析：线段树的成段替换模型，照着写就好了。
代码如下：
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std ;
#define clear( A , X ) memset ( A , X , sizeof A )
#define lson l , m , o << 1
#define rson m + 1 , r , o << 1 | 1
const int maxN = 1000000 ;
int sum[ maxN ] , set [ maxN ] ;
void PushUp ( int o ) {
sum[ o ] = sum[ o << 1 ] + sum[ o << 1 | 1 ] ;
}
void PushDown ( int o , int c ) {
if ( set[ o ] ) {
set[ o << 1 ] = set[ o << 1 | 1 ] = set[ o ] ;
sum[ o << 1 ] = set[ o << 1 ] * ( c - ( c >> 1 ) ) ;
sum[ o << 1 | 1 ] = set[o << 1 | 1 ] * ( c >> 1 ) ;
set[ o ] = 0 ;
}
}
void Build ( int l , int r , int o ) {
set[ o ] = 0;
if ( l == r ) sum[ o ] = 1 ;
else {
int m = ( l + r ) >> 1 ;
Build ( lson ) ;
Build ( rson ) ;
PushUp ( o ) ;
}
}
void Update ( int L , int R , int c , int l , int r , int o ) {
if ( L <= l && r <= R ) {
set[ o ] = c ;
sum[ o ] = c * ( r - l + 1 ) ;
return ;
}
PushDown ( o , r - l + 1 ) ;
int m = ( l + r ) >> 1 ;
if ( L <= m ) Update ( L , R , c , lson ) ;
if ( m < R ) Update ( L , R , c , rson ) ;
PushUp ( o ) ;
}
int work () {
int L , R , c , n , m ;
scanf ( "%d%d" , &n , &m ) ;
Build ( 1 , n , 1 ) ;
for ( int i = 0 ; i < m ; ++ i ) {
scanf ( "%d%d%d" , &L , &R , &c ) ;
Update ( L , R , c , 1 , n , 1 ) ;
}
return sum[ 1 ] ;
}
int main () {
int T , cas ;
for ( scanf ( "%d" , &T ) , cas = 1 ; cas <= T ; ++ cas ) {
printf ( "Case %d: The total value of the hook is %d.\n" , cas , work () ) ;
}
return 0 ;
}