SRM468 - SRM469(1-250pt, 500pt)

SRM 468

DIV1 250pt

题意：给出字典，按照一定要求进行查找。

解法：模拟题，暴力即可。

tag:water

score: 0....

这是第一次AC的代码：

 /*
  * Author:  plum rain
  * score :  0
  */
 #line 11 "T9.cpp"
 #include <sstream>
 #include <stdexcept>
 #include <functional>
 #include <iomanip>
 #include <numeric>
 #include <fstream>
 #include <cctype>
 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <map>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 #define PB push_back
 #define SZ(v) ((int)(v).size())
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<#a<<endl

 typedef vector<string> VS;

 map<char, int> mp;

 class T9
 {
     public:
         string message(vector <string> part, vector <string> dict, vector <string> ke){
             int size = SZ (part);
             for (int i = ; i < size; ++ i)
                 for (int j = ; j < SZ(part[i]); ++ j)
                     mp[part[i][j]] = i+;   

             VS dic; dic.clear();
             size = SZ (dict);
             for (int i = ; i < size; ++ i){
                 string s; s.clear();
                 int n = SZ (dict[i]);
                 for (int j = ; j < n; ++ j)
                     s.PB (mp[dict[i][j]] + '');
                 dic.PB (s);
             }

             for (int i = ; i < size; ++ i)
                 for (int j = ; j < size--i; ++ j){
                     bool ok = false;
                     if (dic[j] < dic[j+]) ok = true;
                     if (dic[j] == dic[j+] && dict[j] > dict[j+])
                         ok = true;
                     if (ok){
                         swap (dic[j], dic[j+]);
                         swap (dict[j], dict[j+]);
                     }
                 }

             string ans; ans.clear();
             size = SZ (ke);
             string s; s.clear();
             for (int i = ; i < size; ++ i)
                 s += ke[i];
             size = SZ (s);
             string tmp; tmp.clear();
             int cnt = ;
             for (int i = ; i < size; ++ i){
                 if (s[i] == '*') cnt += ;
                 if (s[i] == '#') ++ cnt;
                 if (s[i] >= '' && s[i] <= '') tmp.PB(s[i]);
                 if (s[i] == ''){
                     if (tmp.empty() && !cnt){
                         ans.PB (' ');
                         continue;
                     }
                     int pos;
                     for (int j = ; j < SZ (dic); ++ j)
                         if (dic[j] == tmp){
                             pos = j;
                             break;
                         }
                     pos = pos + cnt;
                     ans = ans + dict[pos];
                     cnt = ; tmp.clear();
                     ans.PB (' ');
                 }
             }
             if (tmp.empty() && !cnt) return ans;
             int pos;
             for (int i = ; i < SZ (dic); ++ i)
                 if (dic[i] == tmp){
                     pos = i;
                     break;
                 }
             pos = pos + cnt;
             ans = ans + dict[pos];
             return ans;
         }
 };

这是后来改进的代码：

 /*
  * Author:  plum rain
  * score :  0
  */
 #line 11 "T9.cpp"
 #include <sstream>
 #include <stdexcept>
 #include <functional>
 #include <iomanip>
 #include <numeric>
 #include <fstream>
 #include <cctype>
 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <utility>
 #include <map>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 #define PB push_back
 #define SZ(v) ((int)(v).size())
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<#a<<endl

 typedef vector<string> VS;

 map<char, int> mp;

 bool cmp(pair<string, string> a, pair<string, string> b)
 {
     if (a.first < b.first) return true;
     if (a.first == b.first && a.second < b.second) return true;
     return false;
 }

 class T9
 {
     public:
         string message(vector <string> part, vector <string> dict, vector <string> ke){
             int size = SZ (part);
             mp.clear();
             for (int i = ; i < size; ++ i)
                 for (int j = ; j < SZ(part[i]); ++ j)
                     mp[part[i][j]] = i+;   

             vector<pair<string, string> > D; D.clear();
             size = SZ (dict);
             for (int i = ; i < size; ++ i){
                 string s; s.clear();
                 int n = SZ (dict[i]);
                 for (int j = ; j < n; ++ j)
                     s.PB (mp[dict[i][j]] + '');
                 D.PB (make_pair(s, dict[i]));
             }
             sort (D.begin(), D.end(), cmp);

             size = SZ (ke);
             string s; s.clear();
             for (int i = ; i < size; ++ i)
                 s += ke[i];
             s.PB ('');
             size = SZ (s);

             string tmp; tmp.clear();
             string ans; ans.clear();
             int cnt = ;
             for (int i = ; i < size; ++ i){
                 if (s[i] == '*') cnt += ;
                 if (s[i] == '#') ++ cnt;
                 if (s[i] >= '' && s[i] <= '') tmp.PB(s[i]);
                 if (s[i] == ''){
                     if (tmp.empty() && !cnt){
                         if (i != (size-))
                             ans.PB (' ');
                         continue;
                     }
                     int pos;
                     for (int j = ; j < SZ (D); ++ j)
                         if (D[j].first == tmp){
                             pos = j;
                             break;
                         }
                     pos = pos + cnt;
                     ans = ans + D[pos].second;
                     cnt = ; tmp.clear();
                     if (i != (size-))
                         ans.PB (' ');
                 }
             }
             return ans;
         }
 };
         

SRM 469

DIV1 250pt

题意：给定n * m的网格中，其中有若干点不能被选择，然后从其他点中选出同行且相邻的两点，问有多少种选择方法？(1 <= n,m <= 10^8)

解法：水题

score: 129.09

tag: math, counting

 /*
  * Author:  plum rain
  * score:   129.09
  */
 #line 10 "TheMoviesLevelOneDivOne.cpp"
 #include <iostream>
 #include <cstdio>
 #include <sstream>
 #include <cstring>
 #include <algorithm>
 #include <vector>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 #define SZ(v) ((int)(v).size())
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<#a<<endl

 typedef long long int64;

 struct pos{
     int x, y;
 };

 pos a[];

 bool cmp(pos a, pos b)
 {
     if (a.x < b.x)
         return true;
     if (a.x == b.x && a.y < b.y)
         return true;
     return false;
 }

 int64 gao(int64 size, int64 n, int64 m)
 {
     int i = ;
     int64 tmp = a[].x, dif = ;
     if (a[].y == ) dif = ;
     if (a[].y >= ) dif = ;
     while (i < size){
         if (tmp == a[i].x){
             if (a[i-].y == a[i].y-) ++ dif;
             else dif += ;
             ++ i;
             continue;
         }

         if (a[i-].y == m) -- dif;
         if (a[i].y == ) ++ dif;
         if (a[i].y >= ) dif += ;
         tmp = a[i].x;
         ++ i;
     }
     if (a[i-].y == m) -- dif;

     return n * (m-) - dif;
 }

 class TheMoviesLevelOneDivOne
 {
     public:
         long long find(int n, int m, vector <int> row, vector <int> seat){
             CLR (a);
             int size = SZ (row);
             for (int i = ; i < size; ++ i)
                 a[i].x = row[i], a[i].y = seat[i];
             sort (a, a+size, cmp);

             return (gao(size, n, m));
         }
 };

DIV1 500pt

题意：一个人看电影，该人有一个scare值，并且没看1min电影scare减1。有很多部恐怖电影，每部电影长度不同(length[i])，且每部都有一个瞬间增加scare(s[i])值的时刻。如果scare值小于0，则这个人就会睡着，不能再看电影。请问他安排一个电影观看顺序，使得他能看尽可能多的电影。如果有多组观看顺序看到的电影数相同，则输出字典序最小的。(电影数量小等于20)

解法：如果这道题不要求输出字典序最小的，而是随便输出一组，那就是一道很简单的贪心题。- -

　　　状压DP。假设有n部电影，用opt来表示所有电影是否排进观影序列(1为该电影未进入观影序列)，用num[opt]表示观影序列为opt时所能看的电影数量的最大值，用scare表示观影序列为opt时看完观影序列上的电影后的scare值。则状态转移方程为：

　　　if (opt & (1<<i) && scare >= s[i] && scare >= length[i] - 47) num[opt] = 1 + num[opt - (1<<i)]，for i = 0 to n-1。注意还要记录路径。

score: 0

tag:DP, good

 /*
  * Author:  plum rain
  * score :  0
  */
 #line 10 "TheMoviesLevelTwoDivOne.cpp"
 #include <iostream>
 #include <cstdio>
 #include <sstream>
 #include <cstring>
 #include <algorithm>
 #include <vector>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 #define PB push_back
 #define SZ(v) ((int)(v).size())
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<#a<<endl

 typedef vector<int> VI;

 const int N = ;
 int n;
 vector<pair<int, int> > a;
 int num[<<N], fir[<<N];

 VI DP()
 {
     CLR (num); CLR (fir);
     for (int opt = ; opt < (<<n); ++ opt){
         int flag, tmp = -;
         int sca = ;
         for (int i = ; i < n; ++ i)
             if (!(opt & (<<i))) sca +=  - a[i].first;

         for (int i = ; i < n; ++ i) if (opt & (<<i)){
             int x = ;
             if (sca >= a[i].second && sca >= (a[i].first - ))
                 x =  + num[opt-(<<i)];
             if (x > tmp)
                 tmp = x, flag = i;
         }

         num[opt] = tmp;
         fir[opt] = flag;
     }

     int opt = (<<n) - ;
     VI ret; ret.clear();
     while (opt > ){
         ret.PB (fir[opt]);
         opt -= (<<fir[opt]);
     }
     return ret;
 }

 class TheMoviesLevelTwoDivOne
 {
     public:
         vector <int> find(vector <int> ls, vector <int> sc){
             n = SZ (ls);
             a.clear();
             for (int i = ; i < n; ++ i)
                 a.PB(make_pair(ls[i], sc[i]));

             return (DP());
         }
 };