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【模拟】【set】hdu 4789 ICPC Ranking

时间:2017-03-31 00:08:59      阅读:248      评论:0      收藏:0      [点我收藏+]

标签:putc   cas   using   set   n+1   cto   log   for   opera   

写了一晚上,TLE到死,我选择GG

喵的好像还是前几年我校出的题,这整场都……tm……

改日在战

/*TLE代码*/

#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<map>
#include<algorithm>
#include<iostream>
#include<set>
int e;
using namespace std;
map<string,int>ma;
struct data{
	vector<int>lasts;
	int solved,penalty;
	string name;/*
	data(const int &_solved,const int &_penalty,const vector<int> & _lasts,const string &_name){
		solved=_solved;
		penalty=_penalty;
		lasts=_lasts;
		name=_name;
	}*/
	data(const string &_name){
		solved=penalty=0;
		name=_name;
		lasts.clear();
	}
	data(){}
};
bool operator < (const data &a,const data &b){
	if(a.solved!=b.solved){
		return a.solved>b.solved;
	}
	if(a.penalty!=b.penalty){
		return a.penalty<b.penalty;
	}
	for(int i=a.lasts.size()-1,j=b.lasts.size()-1;i>=0 && j>=0;--i,--j){
		if(a.lasts[i]!=b.lasts[j]){
			return a.lasts[i]<b.lasts[j];
		}
	}
	return a.name>b.name;
}
typedef set<data>::iterator set_ITER;
struct Query{
	string name;
	char pro;
	int t,type;
}Q[50010];
bool operator < (const Query &a,const Query &b){
	if(a.t!=b.t){
		return a.t<b.t;
	}
	return a.type<b.type;
}
set<data>S;
data A[50010],output[50010];
int p;
int B[50010][26]/*,D[50010][26]*/,aftersubs[50010][26];
int C[50010][26],E[50010][26];
int n,m,T,t,ZU;
int main(){
//	freopen("i.in","r",stdin);
	char tmp1[10],tmp2[10];
	scanf("%d",&ZU);
	for(int zu=1;zu<=ZU;++zu){
		printf("Case #%d:\n",zu);
//		memset(aftersubs,0,sizeof(int)*(e+1)*26);
//		memset(B,0,sizeof(int)*(e+1)*26);
//		memset(C,0,sizeof(int)*(e+1)*26);
//		memset(D,0,sizeof(int)*(e+1)*26);
//		memset(E,0,sizeof(int)*(e+1)*26);
		ma.clear();
		e=0;
		S.clear();
		p=0;
		S.insert(data(""));
		scanf("%d%d%d%d",&n,&m,&T,&t);
		for(int i=1;i<=n;++i){
			cin>>Q[i].name;
			if(!ma[Q[i].name]){
				ma[Q[i].name]=++e;
				A[e]=data(Q[i].name);
				S.insert(A[e]);
			}
			scanf("%s%d%s",tmp1,&Q[i].t,tmp2);
			Q[i].pro=tmp1[0];
			if(tmp2[0]==‘E‘){
				Q[i].type=0;
			}
			if(tmp2[0]==‘N‘){
				Q[i].type=1;
			}
			else{
				Q[i].type=2;
			}
		}
		sort(Q+1,Q+n+1);
		for(int i=1;i<=n;++i){
			int id=ma[Q[i].name];
			if(Q[i].t>=t){
				++aftersubs[id][Q[i].pro-‘A‘];
				continue;
			}
			if(Q[i].type==0 || (Q[i].type==2 && C[id][Q[i].pro-‘A‘])){
				continue;
			}
			if(Q[i].type==1){
				++B[id][Q[i].pro-‘A‘];
//				++D[id][Q[i].pro-‘A‘];
			}
			else{
				S.erase(A[id]);
				C[id][Q[i].pro-‘A‘]=Q[i].t;
				E[id][Q[i].pro-‘A‘]=Q[i].t;
				++A[id].solved;
				A[id].penalty+=(Q[i].t+B[id][Q[i].pro-‘A‘]*20);
				A[id].lasts.push_back(Q[i].t);
				S.insert(A[id]);
			}
		}
		int rank=1;
		set_ITER jt=S.begin(); ++jt;
		for(set_ITER it=S.begin();;++it,++rank,++jt){
			if(jt==S.end()){
				break;
			}
			cout<<(*it).name<<‘ ‘<<rank<<‘ ‘<<(*it).solved<<‘ ‘<<(*it).penalty;
			int id=ma[(*it).name];
			for(int i=0;i<m;++i){
				putchar(‘ ‘);
				if(C[id][i]){
					putchar(‘+‘);
					if(B[id][i]){
						printf("%d",B[id][i]);
					}
				}
				else{
					if(aftersubs[id][i]){
						if(B[id][i]){
							printf("-%d",B[id][i]);
						}
						else{
							putchar(‘0‘);
						}
						putchar(‘/‘);
						printf("%d",aftersubs[id][i]);
					}
					else{
						if(B[id][i]){
							printf("-%d",B[id][i]);
						}
						else{
							putchar(‘.‘);
						}
					}
				}
			}
			puts("");
		}
		int From;
		for(int i=1;i<=n;++i){
			if(Q[i].t>=t){
				From=i;
				break;
			}
		}
		for(int i=From;i<=n;++i){
			int id=ma[Q[i].name];
			if(Q[i].type==0 || (Q[i].type==2 && C[id][Q[i].pro-‘A‘])){
				continue;
			}
			else if(Q[i].type==1){
				++B[id][Q[i].pro-‘A‘];
			}
			else{
				C[id][Q[i].pro-‘A‘]=Q[i].t;
				++A[id].solved;
			}
		}
		set_ITER it=S.end(); --it; --it;
		while(1){
			bool allsolved=0;
			while((*it).solved==A[ma[(*it).name]].solved){
				if(it==S.begin()){
					allsolved=1;
					break;
				}
				jt=it; --jt;
				output[++p]=(*it);
				S.erase(it);
				it=jt;
			}
			if(allsolved){
				output[++p]=(*S.begin());
				break;
			}
			data nextFrom;
			if(it!=S.begin()){
				--it;
				nextFrom=(*it);
				++it;
			}
			int id=ma[(*it).name];
			for(int i=0;i<m;++i){
				if(C[id][i]!=E[id][i]){
					E[id][i]=C[id][i];
					data tmp=(*it);
					set_ITER jt=it;
					data __next=*(++jt);
					++tmp.solved;
					tmp.penalty+=(C[id][i]+B[id][i]*20);
					tmp.lasts.push_back(C[id][i]);
					S.erase(it);
					S.insert(tmp);
					it=S.find(tmp);
					++it;
					if((*it).name!=__next.name){
						cout<<tmp.name<<‘ ‘<<(*it).name<<‘ ‘<<tmp.solved<<‘ ‘<<tmp.penalty<<endl;
						it=S.find(nextFrom);
					}
					else{
						--it;
					}
					break;
				}
			}
		}
		for(int i=p,rank=1;i>=1;--i,++rank){
			int id=ma[output[i].name];
			cout<<output[i].name<<‘ ‘<<rank<<‘ ‘<<output[i].solved<<‘ ‘<<output[i].penalty;
			for(int j=0;j<m;++j){
				putchar(‘ ‘);
				if(C[id][j]){
					putchar(‘+‘);
					if(B[id][j]){
						printf("%d",B[id][j]);
					}
				}
				else{
					if(B[id][j]){
						printf("-%d",B[id][j]);
					}
					else{
						putchar(‘.‘);
					}
				}
			}
			puts("");
		}
		for(int i=1;i<=n;++i){
			int id=ma[Q[i].name];
			aftersubs[id][Q[i].pro-‘A‘]=
			B[id][Q[i].pro-‘A‘]=
			C[id][Q[i].pro-‘A‘]=
			E[id][Q[i].pro-‘A‘]=0;
		}
	}
	return 0;
}

【模拟】【set】hdu 4789 ICPC Ranking

标签:putc   cas   using   set   n+1   cto   log   for   opera   

原文地址:http://www.cnblogs.com/autsky-jadek/p/6649269.html

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