【練習12】 貪心演算法 1004 Doing Homework again

演算法思路：貪心。

先對所有的homework排序，deadline大的在後面，然後依次從後往前分配每一天的任務，因為分配給某一天的homework的deadline一定要大於等於這一天的下標（下標依次為1,2,3，...，N），所以遍曆所有可行的（deadline>當前要分配的這一天的下標）homework選擇penalty最大的哪一個就行了，但是選擇之後要記得進行標記，下次遍曆的時候就不要再選了。最後再根據標記統計一下沒有分配的homework的總的penalty就是結果了。

代碼如下，水平有限，寫得有點挫，如果要學習代碼還是參考其他的大牛，不過上面介紹的思路是沒錯的：

//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 Work{int penalty;int deadline;int selected;friend bool operator<(Work w1, Work w2){if(w1.deadline == w2.deadline){return w1.penalty > w2.penalty;}else{return w1.deadline < w2.deadline;}}};#define MAXN 1005Work d[MAXN];int ind[MAXN];int sz;int exits[MAXN];int main(){#ifdef LOCAL//freopen("shuju.txt", "r", stdin);#endifint cas;int N;cin>>cas;for(int cc = 0; cc < cas; cc++){cin>>N;for(int i = 0; i < N; i++){cin>>d[i].deadline;}for(int i = 0; i < N; i++){cin>>d[i].penalty;}for(int i = 0; i < N; i++)//初始化{d[i].selected = 0;}sort(d, d + N);sz = 0;memset(exits, 0, sizeof(exits));memset(ind, 0, sizeof(ind));for(int i = 0; i < N; i++){int temp = d[i].deadline;if(!exits[temp]){ind[temp] = i;sz = max(sz, temp);exits[temp] = 1;}}for(int i = sz; i >= 1; i--){int j = i;while (!exits[j] && j <= sz){j++;}if(j > sz){continue;}int k = ind[j];int xiabiao = k;int m = 0;for(int kk = k; kk < N; kk++){if(!d[kk].selected && m < d[kk].penalty){m = d[kk].penalty;xiabiao = kk;}}if(m != 0){d[xiabiao].selected = 1;}}int ans = 0;for(int i = 0; i < N; i++){if(!d[i].selected){ans += d[i].penalty;}}cout<<ans<<endl;}return 0;}