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| 1 | +#include <algorithm> |
| 2 | +#include <iostream> |
| 3 | +#include <sstream> |
| 4 | +#include <string> |
| 5 | +#include <vector> |
| 6 | +#include <queue> |
| 7 | +#include <set> |
| 8 | +#include <map> |
| 9 | +#include <cstdio> |
| 10 | +#include <cstdlib> |
| 11 | +#include <cctype> |
| 12 | +#include <cmath> |
| 13 | +#include <string> |
| 14 | +#include <cstring> |
| 15 | +using namespace std; |
| 16 | + |
| 17 | +#define REP(i,n) for(int i=0;i<(n);++i) |
| 18 | +#define FOR(i,a,b) for(int i=(a);i<=(b);++i) |
| 19 | +#define RFOR(i,a,b) for(int i=(a);i>=(b);--i) |
| 20 | +#define FOREACH(it,c) for(typeof((c).begin())it=(c).begin();it!=(c).end();++it) |
| 21 | +#define CLR(x) memset((x),0,sizeof((x))) |
| 22 | +#define MP make_pair |
| 23 | +#define MPI make_pair<int, int> |
| 24 | +#define PB push_back |
| 25 | +typedef long long LL; |
| 26 | +typedef vector<int> VI; |
| 27 | +typedef vector<string> VS; |
| 28 | +typedef pair<int, int> PI; |
| 29 | + |
| 30 | +class Solution { |
| 31 | +public: |
| 32 | + string s; |
| 33 | + int len; |
| 34 | + vector<int> mm; |
| 35 | + vector<vector<int> > vv; |
| 36 | + bool isok(int st, int ed) { |
| 37 | + return vv[st][ed] == 1; |
| 38 | + } |
| 39 | + int doit(int idx) { |
| 40 | + int& res = mm[idx]; |
| 41 | + if (res != -1) return res; |
| 42 | + if (idx >= len) return res = 0; |
| 43 | + if (isok(idx, len - 1)) return res = 0; |
| 44 | + res = 1 + doit(idx + 1); |
| 45 | + for (int i = idx + 1; i < len; ++i) { |
| 46 | + if (!isok(idx, i)) continue; |
| 47 | + res = min(res, 1 + doit(i + 1)); |
| 48 | + } |
| 49 | + return res; |
| 50 | + } |
| 51 | + int minCut(string _s) { |
| 52 | + s = _s; |
| 53 | + len = s.length(); |
| 54 | + if (len <= 1) return 0; |
| 55 | + vv.assign(len, vector<int>(len, 0)); |
| 56 | + for (int i = 0; i < len; ++i) vv[i][i] = 1; |
| 57 | + for (int i = 0; i < len - 1; ++i) vv[i][i + 1] = (s[i] == s[i + 1]); |
| 58 | + for (int sz = 2; sz < len; ++sz) { |
| 59 | + for (int i = 0; i < len - sz; ++i) { |
| 60 | + if (s[i] == s[i + sz] && vv[i + 1][i + sz - 1] == 1) vv[i][i + sz] = 1; |
| 61 | + } |
| 62 | + } |
| 63 | + mm.assign(len + 5, -1); |
| 64 | + return doit(0); |
| 65 | + } |
| 66 | +}; |
| 67 | + |
| 68 | +int main() { |
| 69 | + Solution s = Solution(); |
| 70 | + cout << s.minCut("dde") << endl; |
| 71 | + return 0; |
| 72 | +} |
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