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54
Luogu/P1496/P1496.cpp
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54
Luogu/P1496/P1496.cpp
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#include <iostream>
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#include <algorithm>
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#include <vector>
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using std::cin;
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using std::cout;
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const char endl = '\n';
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const int N = 2e4 + 5;
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int n, ans,
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l[N], r[N], // 原位置
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f[N << 1]; // 染色情况(离散化以后),注意开两倍空间(左右端点)
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std::vector<int> nums; // 统计出现的数字,离散化用
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int main() {
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std::ios::sync_with_stdio(false);
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cin.tie(nullptr);
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cin >> n;
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for (int i = 1; i <= n; i++) {
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cin >> l[i] >> r[i];
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// 统计一下哪些数字出现过
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nums.push_back(l[i]);
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nums.push_back(r[i]);
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}
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// 先排序才能用 std::unique 去重,两个函数都在 <algorithm> 里面定义
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std::sort(nums.begin(), nums.end());
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nums.erase(std::unique(nums.begin(), nums.end()), nums.end()); // 去重完移除后面多余的数字
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// 可以讲一下 std::unique 跑完以后数组的特性:前面不重复且单增,后面是多余的数
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for (int i = 1; i <= n; i++) {
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// 求出离散化以后的位置
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int nl = std::lower_bound(nums.begin(), nums.end(), l[i]) - nums.begin(),
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nr = std::lower_bound(nums.begin(), nums.end(), r[i]) - nums.begin();
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// 这里用差分也可以,不过数据范围比较小,这么写也能过
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for (int j = nl; j < nr; j++) {
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f[j] = 1;
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}
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}
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for (int i = 0; i < nums.size() - 1; i++) {
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// 如果被染色就统计一下长度
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if (f[i]) ans += nums[i + 1] - nums[i];
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}
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cout << ans << endl;
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return 0;
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}
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Luogu/P1496/data/P1496_8.in
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Luogu/P1496/data/P1496_8.out
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