hdu 6180 Schedule （贪心）

Schedule

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 456    Accepted Submission(s): 162

Problem Description

There are N schedules, the i-th schedule has start timesi and end time ei (1 <= i <= N). There are some machines. Each two overlapping schedules cannot be performed in the same machine. For each machine the working time is defined as the difference betweentimeend and timestart , where time_{end} is time to turn off the machine and timestart is time to turn on the machine. We assume that the machine cannot be turned off between thetimestart and the timeend.
Print the minimum number K of the machines for performing all schedules, and when only uses K machines, print the minimum sum of all working times.

 

Input

The first line contains an integer T (1 <= T <= 100), the number of test cases. Each case begins with a line containing one integer N (0 < N <= 100000). Each of the next N lines contains two integerssi and ei(0<=si<ei<=1e9).

 

Output

For each test case, print the minimum possible number of machines and the minimum sum of all working times.

 

Sample Input

131 34 62 5

 

Sample Output

2 8

 

题意:

有n个任务，每个任务有起始时间和结束时间。要你用最少的机器，去完成这些任务，一台机器可以用多次，但只能开一次，关一次。并计算在用最少的机器情况下，机器使用时间的总和

解析:

将每个任务的起始时间和结束时间都放在一条时间轴上，每遇到一个任务开始时间就向前找结束时间最近的未被使用的机器，将这个任务的指向这台机器。

ps,不能同C++交，不然会T，得用G++

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <vector>#include <queue>#include <algorithm>using namespace std;typedef long long ll;const int MAX = 100000 + 10;typedef struct {ll val;ll id;int flag;}Point;typedef struct {ll s, e;}MA;typedef struct node {ll val, id;friend bool operator<(node a, node b){return a.val < b.val;}}node;int n;Point num[MAX * 2];int vis[MAX];MA ma[MAX];int machine[MAX];priority_queue<node>q;   //存已被使用过但现在未被使用的机器，按照结束时间排序，越晚结束优先级越高bool cmp(Point p1, Point p2){if (p1.val == p2.val)   //若时间相同，先让结束的时间排在前面即先结束在开始return p1.flag>p2.flag;return p1.val < p2.val;}int main(){int T;scanf("%d", &T);while (T--){while (!q.empty()) q.pop();scanf("%d", &n);for (int i = 1; i <= 2* n; i += 2){ll s, e;scanf("%lld%lld", &s, &e);num[i].val = s;num[i].id = i / 2 + 1;num[i].flag=0;num[i + 1].val = e;num[i + 1].id = i / 2 + 1;num[i+1].flag=1;}sort(num + 1, num + 2 * n + 1, cmp);memset(vis, 0, sizeof(vis));int res = 0;for (int i = 1; i < 2 * n + 1; i++){if (vis[num[i].id] == 0)  //任务的开始{if (q.empty())   //没有机器可用{res++;  //加机器ma[res].s = num[i].val;machine[num[i].id] = res;vis[num[i].id] = 1;}else   //否则取出结束时间最晚的未被使用的机器{node x = q.top();q.pop();machine[num[i].id] = machine[x.id];vis[num[i].id] = 1;}}else if (vis[num[i].id] == 1)  //任务的结束{node a;a.val = num[i].val;a.id = num[i].id;q.push(a);ma[machine[num[i].id]].e = num[i].val;  //更新机器的结束时间}}ll sum = 0;printf("%d ",res);for (int i = 1; i <= res; i++){sum += (ma[i].e - ma[i].s);}printf("%lld\n", sum);}return 0;}