A.Activity planning
题目描述
there is a collection of n activities e={1,2,..,n}, each of which requires the same resource, such as
a lecture venue, etc., and only one activity at a time use this resource. each activity i has a start
time of si and an end time of fi and si<fi. if the activity i is selected, it occupies resources within
the time interval [si,fi). if the interval [si,fi) does not intersect the interval [sj,fj), then the activity i is
said to be compatible with the activity j. that is, when fi<=sj or fj<=si, the activity i is compatible
with the activity j . choose the largest collection of activities that are compatible with each other.
输入格式
the first line is an integer n;
the next n line, two integers per line, si and fi.
输出格式
excluding mutual and compatible maximum active number.
样例输入1
4
1 3
4 6
2 5
1 7
样例输出1
2
数据范围与提示
1<=n<=1000
这道题解法与杭电今年暑假不ac解法相同,采用均摊法
代码实例
#include<stdio.h> struct huodong { int begin; int end; } j[1002]; int main() { int n,i,j,sum,temp; sum = 1; scanf("%d",&n); for(i=0; i<n; i++) scanf("%d %d",&j[i].begin,&j[i].end); for(i=0; i<n-1; i++) { for(j=0; j<n-i-1; j++) { if(j[j].end>j[j+1].end) { temp = j[j].end; j[j].end = j[j+1].end; j[j+1].end = temp; temp = j[j].begin; j[j].begin = j[j+1].begin; j[j+1].begin = temp; } } } temp = j[0].end; for(i=1; i<n; i++) { if(j[i].begin>=temp) { sum++; temp = j[i].end; } } printf("%d\n",sum); return 0; }
上一篇: 爷爷躺着仰游搞笑笑话
下一篇: 谈谈代码重构