线段树 [成段更新] HDU 1698 Just a Hook

成段更新,需要用到延迟标记(或者说懒惰标记),简单来说就是每次更新的时候不要更新到底,用延迟标记使得更新延迟到下次需要更新or询问到的时候.

此处建议在纸上模拟一遍。

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.

 

#include <stdio.h>
#include <iostream>
using namespace std;
const int N = 400000;
int tree[N], flag[N], x, y, value;
void build(int l, int r, int k) {
	tree[k] = 1; //初始为1
	flag[k] = 0;
	if (l == r)
		return;
	int m = (l + r) / 2;
	build(l, m, k * 2); //k*2 即为k的左子树
	build(m + 1, r, k * 2 + 1); // k*2+1 即为k的右子树
	tree[k] = tree[k * 2] + tree[k * 2 + 1]; //更新当前节点的指, 即左子树+右子树
}

//向下更新。 k为更新的节点的，m为更新区间的长度
//将k节点的信息更新到它的左右子树上
void down(int k, int m) {
	if (flag[k]) {
		flag[k * 2] = flag[k * 2 + 1] = flag[k];
		tree[k * 2] = (m - (m / 2)) * flag[k];
		tree[k * 2 + 1] = m / 2 * flag[k];
		flag[k] = 0;
	}
}

void update(int l, int r, int k) {
	if (x <= l && y >= r) {
		flag[k] = value; //存储当前的 value
		tree[k] = (r - l + 1) * value;
		return;
	}
	down(k, r - l + 1); //更新k节点
	int m = (l + r) / 2;
	if (x <= m)
		update(l, m, k * 2);
	if (y > m)
		update(m + 1, r, k * 2 + 1);
	tree[k] = tree[k * 2] + tree[k * 2 + 1];
}

int main() {
	//freopen("in.txt", "r", stdin);
	int T , n , m;
	scanf("%d",&T);
	for (int cas = 1; cas <= T; cas ++) {
		scanf("%d%d",&n,&m);
		build(1 , n , 1);
		while(m--) {
			scanf("%d %d %d", &x, &y, &value);
			update( 1, n ,1);
		}
		printf("Case %d: The total value of the hook is %d.\n",cas , tree[1]);
	}
	return 0;
}