给定一个区间列表interval[] ，其中每个区间包含两个整数L和R ，任务是将区间分配给两个不同的处理器，使得它们对于每个处理器没有重叠的区间。要将 interval[i] 分配给第一个处理器，打印“F”，并将其分配给第二个处理器，打印“S”。
注意：如果没有可能的解决方案，则打印 -1。
例子：
Input: interval[] = {{360, 480}, {420, 540}, {600, 660}}
Output: S, F, S
Explanation:
The intervals assigned to processors are –
Intervals of First Processor {{420, 540}}
Intervals of Second Processor {{360, 480}, {600, 660}}
As there are no overlapping intervals for each processor, it will be a valid solution.
Input: interval[] = {{99, 150}, {1, 100}, {100, 301}, {2, 5}, {150, 250}}
Output: S, F, F, S, S
Explanation:
The intervals assigned to processors are –
Intervals of First Processor {{1, 100}, {100, 301}}
Intervals of Second Processor {{99, 150}, {2, 5}, {150, 250}}
As there are no overlapping intervals for each processor, it will be a valid solution.
方法：想法是使用贪心算法将间隔分配给处理器。
如果处理器的最高结束时间小于或等于某个区间的开始时间，则可以将该区间分配给该处理器。否则，检查另一个处理器。如果无法将任何间隔分配给任何处理器，则没有可能的解决方案。
下面是该方法步骤的说明：
- 在当前的问题中，我们必须根据间隔的顺序进行打印。因此，要保存间隔的顺序，请将间隔与其索引配对。
- 按开始时间对间隔进行排序。即L 。
- 迭代间隔并将间隔分配给处理器，如下所示：
if (interval[i][0] >= firstProcessorEndTime)
answer[interval[i]] = "F"
firstProcessorEndTime =
max(firstProcessorEndTime, interval[i][0])
else if (interval[i][0] >= secondProcessorEndTime)
answer[interval[i]] = "S"
secondProcessorEndTime =
max(secondProcessorEndTime, interval[i][0])
else
print(-1)下面是上述方法的实现：
C++
// C++ implementation for intervals
// scheduling to two processors such
// that there are no overlapping intervals
#include
using namespace std;
// Function to assign the intervals
// to two different processors
void assignIntervals(vector > interval, int n)
{
// Loop to pair the interval
// with their indices
for (int i = 0; i < n; i++)
interval[i].push_back(i);
// sorting the interval by
// their start times
sort(interval.begin(), interval.end());
int firstEndTime = -1;
int secondEndTime = -1;
char fin = ' ';
bool flag = false;
// Loop to iterate over the
// intervals with their start time
for (int i = 0; i < n; i++) {
if (interval[i][0] >= firstEndTime) {
firstEndTime = interval[i][1];
interval[i].push_back('S');
}
else if (interval[i][0] >= secondEndTime) {
secondEndTime = interval[i][1];
interval[i].push_back('F');
}
else {
flag = true;
break;
}
}
// Condition to check if there
// is a possible solution
if (flag)
cout << (-1);
else {
vector form(n, ' ');
for (int i = 0; i < n; i++) {
int indi = interval[i][2];
form[indi] = interval[i][3];
}
// form = ''.join(form)
for (int i = 0; i < form.size(); i++)
cout << form[i] << ",";
}
}
// Driver Code
int main()
{
vector > intervals
= { { 360, 480 }, { 420, 540 }, { 600, 660 } };
// Function Call
assignIntervals(intervals, intervals.size());
return 0;
} Python3
# Python implementation for intervals
# scheduling to two processors such
# that there are no overlapping intervals
# Function to assign the intervals
# to two different processors
def assignIntervals(interval, n):
# Loop to pair the interval
# with their indices
for i in range(n):
interval[i].append(i)
# sorting the interval by
# their startb times
interval.sort(key = lambda x: x[0])
firstEndTime = -1
secondEndTime = -1
fin = ''
flag = False
# Loop to iterate over the
# intervals with their start time
for i in range(n):
if interval[i][0] >= firstEndTime:
firstEndTime = interval[i][1]
interval[i].append('S')
elif interval[i][0] >= secondEndTime:
secondEndTime = interval[i][1]
interval[i].append('F')
else:
flag = True
break
# Condition to check if there
# is a possible solution
if flag:
print(-1)
else:
form = ['']*n
for i in range(n):
indi = interval[i][2]
form[indi] = interval[i][3]
# form = ''.join(form)
print(form, ", ")
# Driver Code
if __name__ == "__main__":
intervals = [[360, 480], [420, 540], [600, 660]]
# Function Call
assignIntervals(intervals, len(intervals))['S', 'F', 'S'] ,时间复杂度： O(NlogN)
辅助空间： O(N)
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