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.

 

 1
10
2
1 5 2
5 9 3 

Sample Output

 Case 1: The total value of the hook is 24.
        

题意：让你修改[l, r]的类型为v, v的种类不同，分数不同，求所有操作后的总和，默认值是1

思路：线段树的区间修改，不算太难

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define lson(x) ((x) << 1)
#define rson(x) ((x) << 1 | 1)
using namespace std;
const int maxn = 100005;

struct seg {
	int w;
	int v;
};

struct segment_tree {
	seg node[maxn<<2];
	
	void update(int pos) {
		node[pos].w = node[lson(pos)].w + node[rson(pos)].w;
	}
	
	void push(int pos, int len) {
		if (node[pos].v > 0) {
			node[lson(pos)].v = node[rson(pos)].v = node[pos].v;
			node[lson(pos)].w = node[pos].v * (len - (len >> 1));
			node[rson(pos)].w = node[pos].v * (len >> 1);
			node[pos].v = 0;
		}
	}

	void build(int l, int r, int pos) {
		node[pos].v = 0;
		if (l == r) {
			node[pos].w = 1;
			return;
		}
		int m = l + r >> 1;
		build(l, m, lson(pos));
		build(m+1, r, rson(pos));
		update(pos);
	}

	void modify(int l, int r, int pos, int x, int y, int z) {
		if (x <= l && y >= r) {
			node[pos].w = (r-l+1) * z;
			node[pos].v = z;
			return;
		}
		push(pos, r-l+1);	
		int m = l + r >> 1;
		if (x <= m)
			modify(l, m, lson(pos), x, y, z);
		if (y > m)
			modify(m+1, r, rson(pos), x, y, z);
		update(pos);
	}
 
	int query(int l, int r, int pos, int x, int y) {
		if (x <= l && y >= r)
			return node[pos].w;
		push(pos, r-l+1);
		int m = l + r >> 1;
		if (x <= m)
			query(l, m, lson(pos), x, y);
		if (y > m)
			query(m+1, r, rson(pos), x, y);
	}
} tree;

int main() {
	int n, q, cas = 1;
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		tree.build(1, n, 1);
		scanf("%d", &q);
		int x, y, z;
		for (int i = 0; i < q; i++) {
			scanf("%d%d%d", &x, &y, &z);
			tree.modify(1, n, 1, x, y, z);
		}
		printf("Case %d: The total value of the hook is %d.\n", cas++, tree.node[1].w);
	}
	return 0;
}