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hdu1698 Just a Hook 线段树：成段替换，总区间求和

2024-07-09 08:44:50   230人阅读

转载请注明出处：http://blog.csdn.net/u012860063

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1698

Problem 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.

题意：题意：有t组测试数据，n为钩子长度（1<=n<=100000）,m为操作的次数。初始时，每个钩子的价值为1，操作由三个数字组成x,y,z表示把区间[x,y]的钩子变成的价值变成z（1代表铜，2银，3金）。

代码如下：

//线段树功能:update:成段替换 (由于只query一次总区间,所以可以直接输出1结点的信息)

#include <cstdio>
#include <algorithm>
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
//lson和rson分辨表示结点的左儿子和右儿子
//rt表示当前子树的根(root),也就是当前所在的结点
const int maxn = 111111;
//maxn是题目给的最大区间,而节点数要开4倍,确切的来说节点数要开大于maxn的最小2x的两倍
int col[maxn<<2];
int sum[maxn<<2];
void PushUp(int rt) //把当前结点的信息更新到父结点
{
	sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void PushDown(int rt,int m)//把当前结点的信息更新给儿子结点
{
	if (col[rt]) 
	{
		col[rt<<1] = col[rt] ;
		col[rt<<1|1] = col[rt];
		sum[rt<<1] = col[rt] * (m - (m >> 1));
		sum[rt<<1|1] = col[rt] * (m >> 1);
		col[rt] = 0;
	}
}
void build(int l,int r,int rt)
{
	col[rt] = 0;
	sum[rt] = 1;//初始化每个节点为1
	if (l == r)
	{
		//scanf("%lld",&sum[rt]);
		return ;
	}
	int m = (l + r) >> 1;
	build(lson);
	build(rson);
	PushUp(rt);
}
void update(int L,int R,int c,int l,int r,int rt) 
{
	if (L <= l && r <= R)
	{
		col[rt] = c;
		sum[rt] = c * (r - l + 1);
		return ;
	}
	PushDown(rt , 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(rt);
}
/*int query(int L,int R,int l,int r,int rt)
{
	if (L <= l && r <= R)
	{
		return sum[rt];
	}
	PushDown(rt , r - l + 1);
	int m = (l + r) >> 1;
	int ret = 0;
	if (L <= m) 
		ret += query(L , R , lson);
	if (m < R)
		ret += query(L , R , rson);
	return ret;
}*/
int main() 
{
	int N , Q ,T , K=0;
	int a , b , c;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&N,&Q);//N为节点数
		build(1 , N , 1);//建树
		while (Q--)//Q为询问次数
		{
		/*	char op[2];
			int a , b , c;
			scanf("%s",op);
			if (op[0] == 'Q') 
			{
				scanf("%d%d",&a,&b);
				printf("%lld\n",query(a , b , 1 , N , 1));
			}
			else
			{
				scanf("%d%d%d",&a,&b,&c);//c为区间a到b增加的值
				update(a , b , c , 1 , N , 1);
			}*/
			scanf("%d%d%d",&a,&b,&c);//c为区间a到b的改变值
			update(a , b , c , 1 , N , 1);
			
		}
		printf("Case %d: The total value of the hook is %d.\n",++K,sum[1]);
	}
	return 0;
}



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