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 stubScanner cin = new Scanner(System.in);int t = cin.nextInt();double[][][][] dp = new double[205][110][1005][2];for(int tt=1; tt<=t; tt++){//initializationint 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 loopdouble 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}