HDU

题意：已知有L件衣服，M个洗衣机，N个烘干机，已知每个机器的工作时间，且每个机器只能同时处理一件衣服，问洗烘完所有衣服所需的最短时间。

分析：

1、优先队列处理出每件衣服最早的洗完时间。

2、优先队列处理出每件衣服最早的烘完时间。

3、用最大的洗完时间与最小的烘完时间相加，取最大值。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
struct Node{
    LL len, et;
    Node(LL l, LL t):len(l), et(t){}
    bool operator < (const Node& rhs)const{
        return et > rhs.et;
    }
};
LL washet[MAXN];
priority_queue<Node> q;
int main(){
    int T;
    scanf("%d", &T);
    int kase = 0;
    while(T--){
        while(!q.empty()) q.pop();
        int L, N, M;
        scanf("%d%d%d", &L, &N, &M);
        LL x;
        for(int i = 0; i < N; ++i){
            scanf("%lld", &x);
            q.push(Node(x, x));
        }
        for(int i = 0; i < L; ++i){
            Node top = q.top();
            q.pop();
            washet[i] = top.et;
            q.push(Node(top.len, top.et + top.len));
        }
        while(!q.empty()) q.pop();
        for(int i = 0; i < M; ++i){
            scanf("%lld", &x);
            q.push(Node(x, x));
        }
        LL ans = 0;
        for(int i = L - 1; i >= 0; --i){
            Node top = q.top();
            q.pop();
            ans = max(ans, washet[i] + top.et);
            q.push(Node(top.len, top.et + top.len));
        }
        printf("Case #%d: %lld
", ++kase, ans);
    }
    return 0;
}