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Re:此题贪心的方法不正确,求大神指点In Reply To:此题贪心的方法不正确,求大神指点 Posted by:qihongqi at 2012-12-21 19:10:41 > 我开始做的贪心的思路是:在所有给定的移动组数中每次选取相容(就是不需要等待的)个数最多的为一次移动,然后再在剩下的组书中选取相容个数最多的为一次移动,。。。直到所有组都移动完,记录移动的次数即可。但是这种贪心选择在每次选相容最多的个数时有重复,这样会影响后面的选择即局部的贪心不是最优的,所以会出错
> 下面是我的代码
> #include <iostream>
> #include<cstring>
> using namespace std;
> class Room
> {
> public:
> int from,to;
> bool isOut;
> Room()
> {
> isOut=false;
> from=0;
> to=0;
> }
> };
> Room room[201];
> void change(Room &r, Room &m)
> {
> int f,t;
> bool tp;
> f=r.from;
> r.from=m.from;
> m.from=f;
> t=r.to;
> r.to=m.to;
> m.to=t;
> tp=r.isOut;
> r.isOut=m.isOut;
> m.isOut=tp;
> }
> void greedySelecter(Room rm[],int len)
> {
> int i,m;
> for(i=0;i<len;i++) if(!rm[i].isOut) {rm[i].isOut=true;break;}
> for(m=i+1;m<len;m++)
> {
> if((rm[m].from-rm[i].to==1)&&rm[m].from%2==0);
> else{
> if(rm[m].from>rm[i].to) { rm[m].isOut=true;i=m;}
> }
> }
> }
> bool check(Room rm[],int n)
> {
> int i;
> for(i=0;i<n;i++) if(!rm[i].isOut) return true;
> return false;
> }
> void mySort(Room rm[],int n)
> {
> int i,j;
> for(i=0;i<n-1;i++)
> for(j=i+1;j<n;j++)
> {
> int k=i;
> if(rm[j].to<rm[i].to) k=j;
> change(rm[i],rm[k]);
> }
> }
> void init(Room rm[],int n)
> {int i;
> for(i=0;i<n;i++)
> {
> rm[i].from=0;
> rm[i].to=0;
> rm[i].isOut=false;
> }
> }
> int main()
> {
> int t;
> cin>>t;
> int n;
> while(t--)
> {
> cin>>n;
> int i;
> int count=0;
> for(i=0;i<n;i++)
> cin>>room[i].from>>room[i].to;
> mySort(room,n);
> while(check(room,n))
> {
> greedySelecter(room,n);
> count++;
> }
> cout<<count*10<<endl;
> init(room,n);
> }
> return 0;
> }
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