题目传送门
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
初始的时候，整个序列都是1，接下来，每次输入l，r，x。表示将l到r之间修改为x且x只会是1、2、3，最后问你序列总和。
题解
- 线段树区间修改模板题
- 维护区间和
AC-Code
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <fstream>
#include <utility>
using namespace std;
typedef long long ll;
typedef pair<int, int> Pii;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);const int INF = 0x7fffffff;
const int MOD = 1e4;
const int MAXN = 2e5 + 7;struct node {
int left;int right;int lazy;int data;
}tree[MAXN << 2];void PushUp(int rt) {
tree[rt].data = tree[rt << 1].data + tree[rt << 1 | 1].data;
}
void PushDown(int rt) {
if (tree[rt].lazy) {
//下传标记tree[rt << 1].lazy = tree[rt].lazy;tree[rt << 1 | 1].lazy = tree[rt].lazy;//更新子节点tree[rt << 1].data = tree[rt << 1].lazy * (tree[rt << 1].right - tree[rt << 1].left + 1);tree[rt << 1 | 1].data = tree[rt << 1 | 1].lazy * (tree[rt << 1 | 1].right - tree[rt << 1 | 1].left + 1);tree[rt].lazy = 0;//清空当前节点标记}
}
void build(int rt, int l, int r) {
tree[rt].lazy = 0; /* ---设置延迟标记--- */tree[rt].left = l;tree[rt].right = r;if (l == r) {
tree[rt].data = 1;return;}int mid = (l + r) >> 1;build(rt << 1, l, mid);build(rt << 1 | 1, mid + 1, r);PushUp(rt);
}
void updata(int rt, int l, int r, int val) {
if (l <= tree[rt].left && tree[rt].right <= r) {
tree[rt].data = val * (tree[rt].right - tree[rt].left + 1);tree[rt].lazy = val;return;}PushDown(rt);int mid = (tree[rt].left + tree[rt].right) >> 1;if (l <= mid) updata(rt << 1, l, r, val);if (r > mid) updata(rt << 1 | 1, l, r, val);PushUp(rt);
}
int Query(int rt, int l, int r) {
if (tree[rt].left >= l && tree[rt].right <= r) {
return tree[rt].data;}PushDown(rt);int ans = 0;int mid = (tree[rt].left + tree[rt].right) >> 1;if (l <= mid) ans += Query(rt << 1, l, r);if (r > mid) ans += Query(rt << 1 | 1, l, r);return ans;
}
int main() {
ios;int T;int zz = 0;while (cin >> T) {
while (T--) {
int n;cin >> n;build(1, 1, n);int m;cin >> m;while (m--) {
int x, y, z;cin >> x >> y >> z;updata(1, x, y, z);}cout << "Case " << ++zz << ": The total value of the hook is " << tree[1].data << "." << endl;}}return 0;
}