离散化
离散化核心思想是将很分散的大数经过pair建立映射,通过二分来查找值。本题是先将所有可能访问到的点存下,再将每个点加多少映射到a数组,最后通过query来二分查找起始点。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
using namespace std;
typedef pair<int, int> PII;
const int N = 300010;
int n, m;
int a[N], s[N];
vector<int> alls;
vector<PII> add, query;
int find(int x)
{
int l = 0, r = alls.size() - 1;
while (l < r)
{
int mid = l + r >> 1;
if (alls[mid] >= x)
r = mid;
else
l = mid + 1;
}
return l + 1;//前缀和下标从1开始,所以要返回l+1
}
vector<int>:: iterator unique(vector<int> &alls)
{
int j = 0;
for (int i = 0; i < alls.size(); i ++)
{
if (i != 0 && alls[i] != alls[i - 1])
{
alls[j++] = alls[i];
}
}
return alls.begin() + j;
}
int main()
{
cin >> n >> m;
for (int i = 0; i < n; i ++)
{
int x, c;
cin >> x >> c;
add.push_back({x, c});
alls.push_back(x);
}
for (int i = 0; i < m; i ++)
{
int l, r;
cin >> l >> r;
query.push_back({l, r});
//alls里加这个是为了防止l和r找不到
alls.push_back(l);
alls.push_back(r);
}
sort(alls.begin(), alls.end());
alls.erase(unique(alls), alls.end());
for (auto item: add)
{
int x = find(item.first);
a[x] += item.second;
}
for (int i = 1; i <= alls.size(); i ++)
{
s[i] = s[i - 1] + a[i];
}
for (auto item : query)
{
int l = find(item.first);
int r = find(item.second);
cout << s[r] - s[l - 1] << endl;
}
}
区间合并
本题也可以用贪心求解。区间合并思路是排序后先确定起始的左端点,然后逐个往后找,如果找到一个区间左端点在当前区间右端点的右边,就把当前区间存下来,把下一个区间再当作起始区间。这有一个细节是在merge函数最后要再加个判断,防止最后一组区间没有被存下。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
using namespace std;
typedef pair<int, int> PII;
vector<PII> segs;
void merge(vector<PII> &segs)
{
vector<PII> t;
int l = -2e9, r = -2e9;
sort(segs.begin(), segs.end());
for (auto item: segs)
{
//合并
if (r >= item.first)
{
r = max(r, item.second);
}
//新区间
else
{
if (l == -2e9)
{
l = item.first;
r = item.second;
}
else
{
t.push_back({l, r});
l = item.first;
r = item.second;
}
}
}
if(l != -2e9) t.push_back({l,r});
segs = t;
}
int main()
{
int n;
cin >> n;
while (n --)
{
int l, r;
cin >> l >> r;
segs.push_back({l, r});
}
merge(segs);
cout << segs.size();
}