在C ++中删除间隔后覆盖的最大点数
在本教程中，我们将讨论一个程序，该程序将在删除间隔后找到最大覆盖点
为此，我们将提供N个间隔和最大范围值。我们的任务是找到一个间隔，该间隔在删除后将为我们提供从1到最大范围值的给定范围内的最大值
示例
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
//找到所需的间隔
void solve(int interval[][2], int N, int Q) {
int Mark[Q] = { 0 };
for (int i = 0; i < N; i++) {
int l = interval[i][0] - 1;
int r = interval[i][1] - 1;
for (int j = l; j <= r; j++)
Mark[j]++;
}
//计算覆盖数字
int count = 0;
for (int i = 0; i < Q; i++) {
if (Mark[i])
count++;
}
int count1[Q] = { 0 };
if (Mark[0] == 1)
count1[0] = 1;
for (int i = 1; i < Q; i++) {
if (Mark[i] == 1)
count1[i] = count1[i - 1] + 1;
else
count1[i] = count1[i - 1];
}
int maxindex;
int maxcoverage = 0;
for (int i = 0; i < N; i++) {
int l = interval[i][0] - 1;
int r = interval[i][1] - 1;
int elem1;
if (l != 0)
elem1 = count1[r] - count1[l - 1];
else
elem1 = count1[r];
if (count - elem1 >= maxcoverage) {
maxcoverage = count - elem1;
maxindex = i;
}
}
cout << "Maximum Coverage is " << maxcoverage << " after removing interval at index " << maxindex;
}
int main() {
int interval[][2] = {
{ 1, 4 },
{ 4, 5 },
{ 5, 6 },
{ 6, 7 },
{ 3, 5 }
};
int N = sizeof(interval) / sizeof(interval[0]);
int Q = 7;
solve(interval, N, Q);
return 0;
}输出结果
Maximum Coverage is 7 after removing interval at index 4