Facebook hacker cup qualification round 题解

除了第三题有点难度之外，整体还是很简单的，满分水过。。。

A 找到第一个#，计算边长，验证一下就行了。其实，20组数据，6分钟内提交，人眼去看也差不多了。。。

int main(){
int t, n;
string str[50];
cin>>t;
for(int test=1; test<=t; test++){
bool result = true;
pair<int, int> pos;
int len = 0;
cin>>n;
for(int i=0; i<n; i++) cin>>str[i];
int row = n;
int col = str[0].size();
bool fir = false;
for(int i=0; i<row; i++){
if(fir==true) break;
for(int j=0; j<col; j++){
if(fir == true) break;
if(str[i][j]=='#'){
pos = make_pair(i, j);
fir = true;
break;
}
}
}
int end = pos.second;
for(end=pos.second; end<col; end++){
if(str[pos.first][end]!='#'){
break;
}
}
len = end-pos.second;
//cout<<"len: "<<len<<endl;
int x = pos.first; int y = pos.second;
for(int i=0; i<row; i++){
for(int j=0; j<col; j++){
if((i-x)<len && (j-y)<len && i>=x && j>=y
&& str[i][j]!='#')
result=false;
if((i-x)>=len || (j-y)>=len || i<x || j<y){
if(str[i][j] == '#')
result = false;
}
if(result == false) break;
}
}
if(result == true)
cout<<"Case #"<<test<<": YES"<<endl;
else cout<<"Case #"<<test<<": NO"<<endl;
}
return 0;
}

B 简单模拟一下就行了，由于对效率要求不高，怎么方便怎么来

int main(){
int t, n, m, p;
// shot, height, name
vector<pair<int, pair<int, string> > > v;
cin>>t;
for(int tt=1; tt<=t; tt++){
cin>>n>>m>>p;
v.clear();
int shot, h;
string name;
for(int i=0; i<n; i++){
cin>>name>>shot>>h;
v.push_back(make_pair(shot, make_pair(h, name)));
}
sort(v.begin(), v.end());
reverse(v.begin(), v.end());
// time, number
vector<pair<int, int> > team[2];
team[0].clear(); team[1].clear();
for(int i=0; i<n; i++){
if(i%2 == 0) team[0].push_back(make_pair(0, i));
else team[1].push_back(make_pair(0, i));
}
vector<string> res;
res.clear();
for(int k=0; k<2; k++){
vector<pair<int, int> > cur, rest;
cur.clear(); rest.clear();
for(int i=0; i<p; i++)
cur.push_back(team[k][i]);
for(int i=p; i<team[k].size(); i++)
rest.push_back(team[k][i]);
for(int i=1; i<=m; i++){
// there is no rest player
if(rest.size() == 0) break;
pair<int, int> outNum, inNum;
for(int j=0; j<cur.size(); j++)
cur[j].first++;
sort(cur.begin(), cur.end(), greater<pair<int, int> >());
sort(rest.begin(), rest.end());
outNum = cur[0]; inNum = rest[0];
cur.erase(cur.begin()); rest.erase(rest.begin());
cur.push_back(inNum);
rest.push_back(outNum);
}
for(int i=0; i<cur.size(); i++){
int index = cur[i].second;
res.push_back(v[index].second.second);
}
}// end outer for loop
sort(res.begin(), res.end());
cout<<"Case #"<<tt<<":";
for(int i=0; i<res.size(); i++)
cout<<" "<<res[i];
cout<<endl;
}
return 0;
}

C 动态规划，关键是注意到给定的概率值只到三位小数，所以状态是有限的。。。

import java.text.DecimalFormat;
import java.util.Scanner;

public class facebook {

/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner cin = new Scanner(System.in);
int t = cin.nextInt();
double[][][][] dp = new double[205][110][1005][2];
for(int tt=1; tt<=t; tt++){
//			initialization
int K = cin.nextInt();
for(int i=0; i<2*K+1; i++)
for(int j=0; j<K+5; j++)
for(int k=0; k<dp[i][j].length; k++)
for(int kk=0; kk<dp[i][j][k].length; kk++)
dp[i][j][k][kk] = 0;

double ps = cin.nextDouble();
double pr = cin.nextDouble();
double oripi = cin.nextDouble();
int pimul = (int) Math.round(oripi*1000);
double pu = cin.nextDouble();
int pumul = (int) Math.round(pu*1000);
double pw = cin.nextDouble();
double pd = cin.nextDouble();
int pdmul = (int) Math.round(pd*1000);
double pI = cin.nextDouble();
//			int pImul = (int) Math.round(pI*1000);
dp[1][1][pimul][1] = oripi*ps+(1.0-oripi)*pr;
dp[1][0][pimul][0] = oripi*(1.0-ps)+(1.0-oripi)*(1.0-pr);
for(int i=2; i<=2*K-1; i++){
int upper = Math.min(K, i);
for(int j=0; j<=upper; j++){
if((i-1-j) >= K) continue;
for(int pp=0; pp<=1000; pp++){
if(j > 0)
dp[i][j][pp][1] = dp[i-1][j-1][pp][1]*(1.0-pw)
+dp[i-1][j-1][pp][0]*(1.0-pI);
dp[i][j][pp][0] = dp[i-1][j][pp][1]*(1.0-pw)
+dp[i-1][j][pp][0]*(1.0-pI);
if(pp==0){
for(int tmp=0; tmp<=pdmul; tmp++){
dp[i][j][pp][0] += dp[i-1][j][pp+tmp][0]*pI;
if(j > 0)
dp[i][j][pp][1] += dp[i-1][j-1][pp+tmp][0]*pI;
}
}
if(pp==1000){
for(int tmp=0; tmp<=pumul; tmp++){
dp[i][j][pp][0] += dp[i-1][j][pp-tmp][1]*pw;
if(j > 0)
dp[i][j][pp][1] += dp[i-1][j-1][pp-tmp][1]*pw;
}
}
if(pp-pumul>=0 && pp!=0 && pp!=1000){
if(j > 0)
dp[i][j][pp][1] += dp[i-1][j-1][pp-pumul][1]*pw;
dp[i][j][pp][0] += dp[i-1][j][pp-pumul][1]*pw;
}
if(pp+pdmul<=1000 && pp!=0 && pp!=1000){
if(j > 0)
dp[i][j][pp][1] += dp[i-1][j-1][pp+pdmul][0]*pI;
dp[i][j][pp][0] += dp[i-1][j][pp+pdmul][0]*pI;
}
double oopi = (double)pp/1000.0;
dp[i][j][pp][1] *= oopi*ps+(1.0-oopi)*pr;
dp[i][j][pp][0] *= oopi*(1.0-ps)+(1.0-oopi)*(1.0-pr);
}// end pp loop
}// end j loop
}// end i loop
double res = 0;
for(int i=K; i<=2*K-1; i++){
for(int pp=0; pp<=1000; pp++){
res += dp[i][K][pp][1];
}
}
DecimalFormat df = new DecimalFormat("0.000000");
System.out.println("Case #"+tt+": "+df.format(res));
}
}// end main method

}