HDU1789_Doing Homework again_贪心、并查集优

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13285    Accepted Submission(s): 7741

Problem Description

Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test. And now we assume that doing everyone homework always takes one day. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.

 

Input

The input contains several test cases. The first line of the input is a single integer T that is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=1000) which indicate the number of homework.. Then 2 lines follow. The first line contains N integers that indicate the deadlines of the subjects, and the next line contains N integers that indicate the reduced scores.

 

Output

For each test case, you should output the smallest total reduced score, one line per test case.

 

Sample Input

333 3 310 5 131 3 16 2 371 4 6 4 2 4 33 2 1 7 6 5 4

 

Sample Output

035

大致题意：

每一项作业都有它的截止期限，逾期未交的减分数。求做作业的方案使减分数最小化。

大体思路：

1)贪心：把所有的作业按照减分数由多到少排序，减分数相同时的按期限由晚到近排序。按排好的顺序依次完成作业。

2)并查集优化。

#include<cstdio>#include<algorithm>struct HW{    int dl,po;    friend operator<(HW& x,HW& y){        return (x.po==y.po)?x.dl>y.dl:x.po>y.po;    }}T[1000];int cas,N,Mdl,Re;int UnionF(int x,int* UF){    int i,j=x;    if(UF[j]!=j){        while(j!=UF[j]) j=UF[j];        while(j!=x)            i=UF[x],UF[x]=j,x=i;    }    return j;}int main(){    //freopen("in.txt","r",stdin);    scanf("%d",&cas);    while(cas--){        Mdl=Re=0;        scanf("%d",&N);        for(int i=0;i<N;i++){            scanf("%d",&T[i].dl);            if(T[i].dl>Mdl) Mdl=T[i].dl;        }        for(int i=0;i<N;i++){            scanf("%d",&T[i].po);            Re+=T[i].po;        }        int UF[Mdl+1];        for(int i=0;i<=Mdl;i++)            UF[i]=i;        std::sort(T,T+N);        for(int i=0;i<N;i++){            int d=UnionF(T[i].dl,UF);            if(d>0){                Re-=T[i].po;                UF[d]=d-1;            }        }        printf("%d\n",Re);    }    return 0;}