【練習12】 貪心演算法 1004 Load Balancing

演算法思路：貪心 + 優先隊列。

jobs需要先對時間排序，再依次選擇time值最大的那個job，把它丟給已耗用時間最少的那個server（通過優先隊列尋找）。這個過程中要記錄每次配對的job和server編號，最後一次性輸出。

幾天沒有寫代碼，代碼寫得很挫，WA了無數遍，o(╯□╰)o coding是個偉大的工程，欲速則不達，還是十年磨一劍吧！

代碼如下：

//模板開始#include <string>   #include <vector>   #include <algorithm>   #include <iostream>   #include <sstream>   #include <fstream>   #include <map>   #include <set>   #include <cstdio>   #include <cmath>   #include <cstdlib>   #include <ctime>#include <iomanip>#include <queue>#include <string.h>#define SZ(x) (int(x.size()))using namespace std;int toInt(string s){istringstream sin(s); int t; sin>>t; return t;}template<class T> string toString(T x){ostringstream sout; sout<<x; return sout.str();}typedef long long int64;int64 toInt64(string s){istringstream sin(s); int64 t; sin>>t;return t;}template<class T> T gcd(T a, T b){ if(a<0) return gcd(-a, b);if(b<0) return gcd(a, -b);return (b == 0)? a : gcd(b, a % b);}#define LOCAL//模板結束（通用部分）struct Server{int priority;int server_id;friend bool operator<(Server n1, Server n2){return n1.priority > n2.priority;}};struct Job{int job_id;int time;friend bool operator<(Job a, Job b){return a.time > b.time;}};#define MAXN 100005int cas;int N, M;priority_queue<Server> pq;Server temp;int d[MAXN];Job e[MAXN];int f[MAXN];void init(){while(!pq.empty()){pq.pop();}for(int i = 0; i < M; i++){temp.server_id = i;temp.priority = 0;pq.push(temp);}}int main(){#ifdef LOCAL//freopen("shuju.txt", "r", stdin);#endifint a;//cin>>cas;scanf("%d", &cas);for(int cc = 0; cc < cas; cc++){//cin>>N>>M;scanf("%d%d", &N, &M);init();cout<<N<<endl;for(int i = 0; i < N; i++){//cin>>a;scanf("%d", &e[i].time);e[i].job_id = i;}sort(e, e + N);/*cout<<endl;for(int i = 0; i < N; i++){cout<<d[i]<<" ";}cout<<endl;*/for(int i = 0; i < N; i++){temp = pq.top();pq.pop();temp.priority += e[i].time;pq.push(temp);int b = temp.server_id;int c = e[i].job_id;f[c] = b;}for(int i = 0; i < N; i++){if(i == N - 1){cout<<f[i];}else{cout<<f[i]<<" ";}}cout<<endl;}return 0;}