TCO 2014 Round 1A
顺利搞出 A B 两题，然后压线晋级了，手速场。
A 题 ， 求排列最小的，肯定从后往前来做，维护一个最小的set，只是第一个字母要特判一下。
1: #line 5 "EllysSortingTrimmer.cpp"
2: #include <vector>
3: #include <list>
4: #include <map>
5: #include <set>
6: #include <deque>
7: #include <stack>
8: #include <bitset>
9: #include <algorithm>
10: #include <functional>
11: #include <numeric>
12: #include <utility>
13: #include <sstream>
14: #include <iostream>
15: #include <iomanip>
16: #include <cstdio>
17: #include <cmath>
18: #include <cstdlib>
19: #include <ctime>
20: #include <cstring>
21: using namespace std;
22: int i,j,k;
23: #define REP(i,n) for((i)=0;(i)<(int)(n);(i)++)
24: #define snuke(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
25:
26:
27: class EllysSortingTrimmer
28: {
29: public:
30: string getMin(string S, int L)
31: {
32: vector<char> s;
33: for(int i=1; i<S.size(); i++) s.push_back(S[i]);
34: sort(s.begin(),s.end());
35: vector<char> t;
36: for(int i=0; i<L-1; i++) t.push_back(s[i]);
37: t.push_back(S[0]);
38: sort(t.begin(), t.end());
39: string ret;
40: for(int i=0; i<L; i++) ret += t[i];
41: return ret;
42: }
43:
44:
45: };
46:
B题 不要搞成匹配的问题，每一个字母都有一个位置范围，所以只要维护一个set，保证其能够被写入，其余时间都插入最小就OK
1: #line 5 "EllysScrabble.cpp"
2: #include <vector>
3: #include <list>
4: #include <map>
5: #include <set>
6: #include <deque>
7: #include <stack>
8: #include <bitset>
9: #include <algorithm>
10: #include <functional>
11: #include <numeric>
12: #include <utility>
13: #include <sstream>
14: #include <iostream>
15: #include <iomanip>
16: #include <cstdio>
17: #include <cmath>
18: #include <cstdlib>
19: #include <ctime>
20: #include <cstring>
21: using namespace std;
22: int i,j,k;
23: #define REP(i,n) for((i)=0;(i)<(int)(n);(i)++)
24: #define snuke(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
25:
26:
27: class EllysScrabble
28: {
29: public:
30: string getMin(string letters, int maxDistance)
31: {
32: pair<char, int> X;
33: vector<pair<char, int> > S;
34: for(int i=0; i<letters.size(); i++)
35: {
36: S.push_back( make_pair(letters[i], i));
37: }
38:
39: string ret;
40:
41: vector<pair<char, int> > tmp;
42: for(int i=0; i< min(letters.size() *1.0,maxDistance + 1.0); i++) tmp.push_back(make_pair(letters[i], i));
43: for(int i=0; i<letters.size(); i++)
44: {
45: // search
46: sort(tmp.begin(), tmp.end());
47: bool flag = 0;
48: for(int j=0; j< tmp.size(); j++)
49: {
50: if(i - tmp[j].second == maxDistance)
51: {
52: ret += tmp[j].first;
53: tmp.erase(tmp.begin() + j);
54: flag = true;
55: break;
56: }
57: }
58: if(flag == false)
59: {
60: ret += tmp[0].first;
61: tmp.erase(tmp.begin());
62: }
63:
64: // push
65: if(i + maxDistance + 1 < letters.size())
66: {
67: tmp.push_back(S[ i+ maxDistance +1]);
68: }
69: }
70: return ret;
71: }
72:
73:
74: };
75:
C题 这个题目开始的时候没搞出来，其实当时已经发下了 YYY， NY YN3种情况肯定会有一个是亮的。但是没有仔细想。这个为什么贪心是对的呢？给出每一个序列，NY YN YYY一定是最大的，这个仔细推了一下好像是对的！但是有没有严格证明出来。
#line 5 "EllysLamps.cpp" #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <ctime> #include <cstring> using namespace std; int i,j,k; #define REP(i,n) for((i)=0;(i)<(int)(n);(i)++) #define snuke(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++) class EllysLamps { public: int getMin(string lamps) { int num = 0; for(int i=0; i<lamps.size();) { if(i+1 < lamps.size() && lamps[i] != lamps[i+1]) { i+=2; num++; }else if(i+2 < lamps.size() && lamps[i] == lamps[i+1] && lamps[i] == lamps[i+2] && lamps[i]== 'Y') { i+=3; num++; }else i++; } return num; } };