zoukankan      html  css  js  c++  java
  • hdu 3288 Resource Allocation

    题目连接

    http://acm.hdu.edu.cn/showproblem.php?pid=3288

    Resource Allocation

    Description

    HDU-Sailormoon is made up of three girls~~~wj, xq, lff, usually they work together ---- solve a variety of problems. So they has to use some resources in the process.
    In order to make it more convenient, they will put some resources in a box big enough, each resource has its ID and level of priority. When they want a kind of resources, they will give its ID and get it from the box. If there are several ones available in the box, they will get the highest priority ones. If there are still several ones available, they will get the one which puts in the box first.

    Input

    The input will consist of several cases, please deal with till the end of file. Each case contains a integer N(0<N<=10000), representing there are N steps following. For example, if input is "R x y"(x, y are integers,0<=x,y<=10000), representing they put a resource to the box, its ID is x, and its priority is y(the higher the priority is, the smaller the y is). If input is "name r" (name may be "wj" or "xq" or "lff", r is an integer,0<=r<=10000), representing one girl called "name" wants a resource, which ID is r.

    Output

    When the input is "R x y", the resource will mark a number k (begin from 1). When the input is "name r", please find out a resource in the box, if there is one available, print "name gets Num k: x y!", name referred to the input, k is the mark number of resource, x is the resource's ID and y is the level of priority, or print "No one fits!".

    Sample Input

    9
    R 1 5
    R 2 3
    R 1 5
    R 2 0
    wj 1
    xq 2
    lff 3
    lff 2
    xq 2

    Sample Output

    wj gets Num 1: 1 5!
    xq gets Num 4: 2 0!
    No one fits!
    lff gets Num 2: 2 3!
    No one fits!

    优先队列。。

     1 #include<algorithm>
     2 #include<iostream>
     3 #include<cstdlib>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<vector>
     7 #include<queue>
     8 #include<set>
     9 using std::cin;
    10 using std::cout;
    11 using std::endl;
    12 using std::find;
    13 using std::sort;
    14 using std::set;
    15 using std::pair;
    16 using std::vector;
    17 using std::multiset;
    18 using std::priority_queue;
    19 #define pb(e) push_back(e)
    20 #define sz(c) (int)(c).size()
    21 #define mp(a, b) make_pair(a, b)
    22 #define all(c) (c).begin(), (c).end()
    23 #define iter(c) decltype((c).begin())
    24 #define cls(arr,val) memset(arr,val,sizeof(arr))
    25 #define cpresent(c, e) (find(all(c), (e)) != (c).end())
    26 #define rep(i, n) for (int i = 0; i < (int)(n); i++)
    27 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)
    28 const int N = 10010;
    29 typedef unsigned long long ull;
    30 struct Node {
    31     int fix, pos;
    32     Node(int i = 0, int j = 0) :fix(i), pos(j) {}
    33     inline friend bool operator<(const Node &a, const Node &b) {
    34         return a.fix == b.fix ? a.pos > b.pos : a.fix > b.fix;
    35     }
    36 };
    37 priority_queue<Node> que[N];
    38 int main() {
    39 #ifdef LOCAL
    40     freopen("in.txt", "r", stdin);
    41     freopen("out.txt", "w+", stdout);
    42 #endif
    43     char buf[10];
    44     int n, id, fix;
    45     while (~scanf("%d", &n)) {
    46         int pos = 1;
    47         rep(i, n) {
    48             scanf("%s", buf);
    49             if (buf[0] == 'R') {
    50                 scanf("%d %d", &id, &fix);
    51                 que[id].push(Node(fix, pos++));
    52             } else {
    53                 scanf("%d", &id);
    54                 if (que[id].empty()) { puts("No one fits!"); continue; }
    55                 Node t = que[id].top(); que[id].pop();
    56                 printf("%s gets Num %d: %d %d!
    ", buf, t.pos, id, t.fix);
    57             }
    58         }
    59         rep(i, N) while (!que[i].empty()) que[i].pop();
    60      }
    61     return 0;
    62 }
    View Code
    By: GadyPu 博客地址:http://www.cnblogs.com/GadyPu/ 转载请说明
  • 相关阅读:
    Android 重写系统Crash处理类,保存Crash信息到SD卡 和 完美退出程序的方法
    LeetCode第二十四题-交换链表中节点值
    LeetCode第二十三题-合并n个有序链表
    LeetCode第二十二题-创建n对括号
    LeetCode第二十一题-对两个有序链表排序
    LeetCode第二十题-有效的括号
    LeetCode第十九题-链表节点的删除
    LeetCode第十八题-四数之和
    LeetCode第十七题-电话号码的字母组合
    LeetCode第十六题-找出数组中三数之和最接近目标值的答案
  • 原文地址:https://www.cnblogs.com/GadyPu/p/4619363.html
Copyright © 2011-2022 走看看