[DP] arrange the schedule

Question link: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemcode = 3538.

Question: for example, question.

Question: assume a group of data... (N1) .... (N2) .... (N2) because the three parts are independent events, the total number is multiplied by the three parts. (1) (3) easy to find, that is, 3 ^ N1, 3 ^ N3. For (2), when the head and tail are the same (as in the example) an = 2an-1 + 3an-2 can be introduced; (a1 = 0, a2 = 3) then an = (3 (-1) ^ (n-2) + 3 ^ N)/4; when the head and tail do not want the same (such as .... B) An = 2an-1 + 3an-2; (a1 = 1; a2 = 2) then an = (-1) ^ (n-1) + 3 ^ N) /4 (The calculation is 4 to retrieve the reverse element );

 

Note that when m = 0, and the adjacent two letters are the same (the result is 0 ).

 

 1 /***Good Luck***/ 2 #define _CRT_SECURE_NO_WARNINGS 3 #include <iostream> 4 #include <cstdio> 5 #include <cstdlib> 6 #include <cstring> 7 #include <string> 8 #include <algorithm> 9 #include <stack>10 #include <map>11 #include <queue>12 #include <vector>13 #include <set>14 #include <functional>15 #include <cmath>16 #include <numeric>17 18 #define Zero(a) memset(a, 0, sizeof(a))19 #define Neg(a)  memset(a, -1, sizeof(a))20 #define All(a) a.begin(), a.end()21 #define PB push_back22 #define inf 0x3f3f3f3f23 #define inf2 0x7fffffffffffffff24 #define ll long long25 using namespace std;26 //#pragma comment(linker, "/STACK:102400000,102400000")27 void get_val(int &a) {28     int value = 0, s = 1;29     char c;30     while ((c = getchar()) == ‘ ‘ || c == ‘\n‘);31     if (c == ‘-‘) s = -s; else value = c - 48;32     while ((c = getchar()) >= ‘0‘ && c <= ‘9‘)33         value = value * 10 + c - 48;34     a = s * value;35 }36 const int maxn = 14;37 pair<char, int> pr[maxn];38 int n, m;39 ll mod = 1000000007;40 inline ll modxp(ll a, ll b, ll mod) {41     ll ret = 1;42     ll tmp = a;43 44     while (b) {45         if (b & 1) ret = ret* tmp %mod;46         tmp = tmp * tmp % mod;47         b >>= 1;48     }49     return ret;50 }51 52 bool cmp(pair<char, int> a, pair<char, int> b) {53     return a.second < b.second;54 }55 int main() {56     //freopen("data.out", "w", stdout);57     //freopen("data.in", "r", stdin);58     //cin.sync_with_stdio(false);59     while (cin >> n >> m) {60         ll ans = 1;61         if (m == 0) {62             cout << (4 * modxp(3, n - 1, mod)) % mod << endl;63             continue;64         }65         for (int i = 1; i <= m; ++i) {66             cin >> pr[i].second >> pr[i].first;67         }68         sort(pr + 1, pr + m + 1, cmp);69         ans *= modxp(3, pr[1].second - 1, mod);70         ans *= modxp(3, n - pr[m].second, mod);71         ans %= mod;72         ll t;73         bool flag = false;74         for (int i = 2; i <= m; ++i) {75             if (pr[i].first == pr[i - 1].first && pr[i].second - pr[i - 1].second == 1) {76                 cout << 0 << endl;77                 flag = true;78                 break;79             }80             if (pr[i].second - pr[i - 1].second == 1) continue;81 82             else {83                 t = pr[i].second - pr[i - 1].second;84                 if (pr[i].first == pr[i - 1].first) {85                     ans *= ((3 * ((t & 1) ? -1 : 1) + modxp(3, t, mod))  * 250000002) % mod ;86                     ans %= mod;87                 }88                 else {89                     ans *= ((((t & 1) ? 1 : -1) + modxp(3, t, mod)) * 250000002) % mod;90                     ans %= mod;91                 }92             }93         }94         if (!flag) 95         cout << ans << endl;96     }97     return 0;98 }

 

[DP] arrange the schedule