vector的小小应用，大大功能

Description

As a contestant, you must be familiar with the rules of ACM-ICPC. Teams are ranked according to the most problems solved. Teams who solve the same number of problems are ranked by least total time. The total time is the sum of the time consumed for each problem solved. The time consumed for a solved problem is the time elapsed from the beginning of the contest to the submittal of the accepted run plus 20 penalty minutes for every rejected run for that problem regardless of submittal time. There is no time consumed for a problem that is not solved. Teams who solve the same number of problems and the same total time are ranked by their names in alphabetical order. 

Any submit of a problem by those teams who has solved this problem before would be ignored!

Caspar (a coach of SCPC) wants to know when the rank of his students’ team changed. He makes a curve to show the rank’s changes of his students’ team. Now, give you the information of every submission. Your problem is to calculate the changing points of a giving team. 

Input

 

There is ONLY ONE TESTCASE.

The first line is an integer N (1<=N<=200) which indicates the number of submissions, and the number of teams is less than 20.

Each of the following N lines has four operations A (indicating the Name of the team who submitted this code, no more than 20 characters), B(an integer indicating the submitted time), C (indicating the ID of problem, only one uppercase character), D (indicating the result of this run, ‘Y’ means accepted and ‘N’ means not accepted).

Then an integer M followed in one line.

Each of the following M lines has only one string indicating the team name which Caspar wants to check its curve.

We promise that there is at most one submit in the same time and each of those teams Caspar want to check has at least one submit in this competition.

Output

For each team that Caspar wants to check its curve. You should output the changing points of the team in several lines. Each line has two number T, R indicating the time when the rank has been changed and rank. You should output as this format “time:T rank:R”. There is a blank line followed.

After that, you should output the final rank of all teams.

Sample Input

16

TEAM1 1 A Y

TEAM2 2 A N

TEAM2 3 A Y

Dream 4 A Y

TEAM3 5 A Y

TEAM3 6 B Y

TEAM4 7 A Y

TEAM4 8 B Y

TEAM4 9 C Y

Dream 10 B Y

Dream 11 C Y

TEAM3 12 C Y

TEAM1 13 B Y

TEAM1 14 C Y

TEAM1 15 D Y

Dream 16 D Y

2

Dream

TEAM1

Sample Output

time:4 rank:2

time:6 rank:3

time:8 rank:4

time:10 rank:3

time:11 rank:2

time:12 rank:3

time:15 rank:4

time:16 rank:1



time:1 rank:1

time:6 rank:2

time:8 rank:3

time:10 rank:4

time:15 rank:1

time:16 rank:2



Dream

TEAM1

TEAM3

TEAM4

TEAM2

没有用vector的做法，采用的是另外的两个二维数组

一个用来记时间，一个用来记排名。此外，还用了一个一

维数组，来统计有多少次改动。

然而，所有这些，用一个vector就可以搞定了。

先贴出没有用vector的代码。

#include "iostream"
#include "cstring"
#include "algorithm"
#include "string"
#include "map"
using namespace std;
#define sz 21
#define ca 201
int tm[sz][ca], rk[sz][ca], wg[sz][ca];
bool AC[sz][ca];
struct submit
{
    string na;
    int ti;
    char pr, sb;
}s[ca];
struct team
{
    string na;
    int an, at, ra;
}t[sz], tt[sz];
bool check(team a, team b)
{
    if(a.an > b.an) return true;
    if(a.an==b.an && a.at<b.at) return true;
    if(a.an==b.an && a.at==b.at && a.na<b.na) return true;
    return false;
}
int main()
{
    int n, k1 = 0, k2 = 0, flag1, flag2, df[sz], who, whi, nowtime;
    string ne;
    map<string, int> nm; //name
    map<char, int> pm; //probelm
    for(int i=0; i<21; i++)
    {
        t[i].an = 0; t[i].at = 0;
        t[i].ra = 0; df[i] = 0;
    }
    memset(tm, 0, sizeof(tm)); memset(rk, 0, sizeof(rk));
    memset(wg, 0, sizeof(wg)); memset(AC, false, sizeof(AC));
    cin>>n;
    for(int i=0; i<n; i++)
    {
        flag1 = 1; flag2=1;
        cin>>s[i].na>>s[i].ti>>s[i].pr>>s[i].sb;
        nowtime = s[i].ti;
        for(int j=0; j<i; j++)
        {
            if(s[i].na==s[j].na) flag1=0;
            if(s[i].pr==s[j].pr) flag2=0;
        }
        if(flag1) 
        {
            t[k1].na = s[i].na; //存入姓名
            nm[s[i].na]=k1; k1++;
        }
        if(flag2)
        {
            pm[s[i].pr]=k2;
            k2++;
        }
        who = nm[s[i].na]; whi = pm[s[i].pr];
        if(s[i].sb=='N') wg[who][whi]++;
        else if(AC[who][whi]==false)
        {
            AC[who][whi]=true;
            t[who].an++; 
            t[who].at+=(nowtime+20*wg[who][whi]);
        }
        for(int i=0; i<k1; i++)
            tt[i] = t[i];
        sort(tt, tt+k1, check);
        for(int j=0; j<k1; j++)
        {
            who = nm[tt[j].na]; 
            if(t[who].ra!=(j+1)) //j+1是现在的排名
            {
                t[who].ra = j+1; df[who]++;
                tm[who][df[who]]=nowtime; //记录提交时的时间
                rk[who][df[who]]=j+1;
            }
        }
    }
    cin>>n;
    while(n--)
    {
        cin>>ne;
        for(int i=1; i<=df[nm[ne]]; i++)
            cout<<"time:"<<tm[nm[ne]][i]<<" rank:"<<rk[nm[ne]][i]<<endl;
        cout<<endl;
    }
    for(int i=0; i<k1; i++)
        cout<<tt[i].na<<endl;
}

下面给出用了vector的代码。

#include <string>
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
struct Info
{
    int iTime;
    int iRank;
};
struct Team
{
    int Problems[36];  //小于0表示已经AC，大于0表示WA次数，等于0表示还没提交
    int AcCount;
    int total_time;
    string name;
    int rank;
    vector<Info> curve; //将结构体Info放入vector中，这个用法比较新
};
bool Cmp(Team t1,Team t2)
{
    if(t1.AcCount > t2.AcCount)return true;
    if(t1.AcCount < t2.AcCount)return false;
    if(t1.total_time < t2.total_time)return true;
    if(t1.total_time > t2.total_time)return false;
    return t1.name < t2.name;
}
int N,M;
Team Teams[20];
int Team_Count = 0;
int GetID(string str) //将字符串映射到数字，便于用计算机处理，其实可以用map来实现
{
    int i;
    for(i=0;i<Team_Count;i++)
        if(str == Teams[i].name) return i;//已经存在了，就返回他的number
    Teams[Team_Count].name = str;  //不存在，加入Teams中
    Team_Count++;
    return Team_Count-1;
}

void Init()
{
    int i,j;
    for(i=0;i<20;i++)
    {
        for(j=0;j<36;j++)
            Teams[i].Problems[j] = 0;
        Teams[i].AcCount = 0;
        Teams[i].name = "";
        Teams[i].rank = 21;
        Teams[i].total_time = 0;
    }
}
int main()
{
    Init();
    int i,j, NID, PID;
    string teamName;
    char results,ProblemID;
    int SubTime;
    cin>>N;
    for(i=1;i<=N;i++)
    {
        cin>>teamName>>SubTime>>ProblemID>>results;
        NID = GetID(teamName); PID = ProblemID-'A';
        if( Teams[NID].Problems[PID] < 0 )continue;  //忽略已经AC的题目
        if(results == 'Y')
        {
            Teams[NID].total_time += (SubTime+20*Teams[NID].Problems[PID]);
            Teams[NID].Problems[PID] = -1;
            Teams[NID].AcCount++;
        }
        else
            Teams[NID].Problems[PID]++;
        sort(Teams,Teams+Team_Count,Cmp);
        for(j=0;j<Team_Count;j++)
        {
            if(Teams[j].rank != j)
            {
                Info TMP;
                TMP.iRank = 1+j;
                TMP.iTime = SubTime;
                Teams[j].rank = j;
                Teams[j].curve.push_back(TMP); //将变化的排名存入Teams[j]的vector中
            }
        }
    }
    cin>>M;
    while(M--)
    {
        cin>>teamName;
        int xb = GetID(teamName);
        for(i=0;i<Teams[xb].curve.size();i++)
            cout<<"time:"<<Teams[xb].curve[i].iTime<<" rank:"<<Teams[xb].curve[i].iRank<<endl;
        cout<<endl;
    }
    for(i=0;i<Team_Count;i++)
        cout<<Teams[i].name<<endl;
}

总结：

在一个结构体中放入一个vector来可以统计结构体

以外的内容，省去了用多个数组才能实现的麻烦。