题目地址 :
http://acm.hdu.edu.cn/showproblem.php?pid=1698
题目描述 :
Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3841 Accepted Submission(s): 1675
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.
标准的线段树, 成段更新 ,...... 具体看 代码 注释 .
代码如下 :
/*
Coded By : MiYu
Link : http://www.cnblogs.com/MiYu || http://www.cppblog.com/MiYu
Author By : MiYu
Test : 1
Program : 1698
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
using namespace std;
typedef struct seg_tree{
int left, right, col;
bool cov; //标记当前线段是否被覆盖, 如果true, 表示这一段线段的值都为 col. false则相反
int mid (){ return (left + right) >> 1; }
}SEG;
SEG seg[300010];
void creat ( int beg, int end, int rt = 1 ){
seg[rt].left = beg;
seg[rt].right = end;
seg[rt].col = 1;
seg[rt].cov = true;
if ( beg == end ) return;
int mid = seg[rt].mid();
creat ( beg, mid, rt << 1 );
creat ( mid + 1, end, ( rt << 1 ) + 1 );
}
void modify ( int beg, int end, int val, int rt = 1 ){
int LL = rt << 1;
int RR = ( rt << 1 ) + 1;
if ( seg[rt].left == beg && seg[rt].right == end ){ //线段被覆盖, 标记 cov 为true
seg[rt].cov = true;
seg[rt].col = val;
return ;
}
if ( seg[rt].cov ){ //如果线段曾经被覆盖, 标记 false, 将col往下传
seg[rt].cov = false;
seg[LL].col = seg[RR].col = seg[rt].col;
seg[LL].cov = seg[RR].cov = true;
}
int mid = seg[rt].mid();
if ( end <= mid ){
modify ( beg, end, val, LL );
} else if ( beg > mid ) {
modify ( beg, end, val, RR );
} else {
modify ( beg, mid, val, LL );
modify ( mid + 1, end, val, RR );
}
}
int quy ( int beg, int end, int rt = 1 ){
if ( seg[rt].cov ){ // 线段如果是被覆盖的 , 直接返回这一段区间的值
return ( seg[rt].right - seg[rt].left + 1 ) * seg[rt].col;
}
int mid = seg[rt].mid();
return quy ( beg, mid, rt << 1 ) + quy ( mid + 1, end, ( rt << 1 ) + 1 );
}
int main ()
{
int T, ca = 1;
scanf ( "%d", &T );
while ( T -- ){
int N;
scanf ( "%d", &N );
creat ( 1, N );
int M;
scanf ( "%d", &M );
for ( int i = 1; i <= M; ++ i ){
int beg, end, val;
scanf ( "%d%d%d", &beg, &end, &val );
modify ( beg, end, val );
}
printf ( "Case %d: The total value of the hook is %d.\n", ca++,quy( 1, N ) );
}
return 0;
}
/*
1
10
2
1 5 2
5 9 3
*/
/* 此为一牛人代码 , 速度 非常快 !!!! ０ｒｚ．．．．．．．．．
#include<stdio.h>
int a[100001][3],c[100001];
int main()
{
int t,i,j,n,m,sum,v,w=1;
scanf("%d",&t);
while(t--&&scanf("%d %d",&n,&m))
{
sum=0;
for(i=1;i<=m;i++)
scanf("%d %d %d",&a[i][0],&a[i][1],&a[i][2]);
for(i=1;i<=n;i++)
{
v=1;
for(j=m;j>=1;j--)
{
if(a[j][0]<=i&&a[j][1]>=i)
{
v=a[j][2];
break;
}
}
sum+=v;
}
printf("Case %d: The total value of the hook is %d.\n",w++,sum);
}
}
*/
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