标签:字典序 ble fine ack memset unsigned return tac ons
题意:有n个字符串,只能将其逆转,不能交换位置,且已知逆转某字符串需要消耗的能量,问将这n个字符串按字典序从小到大排序所需消耗的最少能量。
分析:每个字符串要么逆转,要么不逆转,相邻两个字符串进行比较,从而可得4个状态转移方程。
#include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define lowbit(x) (x & (-x)) const double eps = 1e-8; inline int dcmp(double a, double b){ if(fabs(a - b) < eps) return 0; return a > b ? 1 : -1; } typedef long long LL; typedef unsigned long long ULL; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const LL LL_INF = 0x3f3f3f3f3f3f3f3f; const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const int MAXN = 100000 + 10; const int MAXT = 10000 + 10; using namespace std; LL c[MAXN]; string s[MAXN][2]; LL dp[MAXN][2]; int main(){ int n; scanf("%d", &n); for(int i = 0; i < n; ++i){ scanf("%I64d", &c[i]); } string tmp; for(int i = 0; i < n; ++i){ cin >> tmp; s[i][0] = tmp; reverse(tmp.begin(), tmp.end()); s[i][1] = tmp; } memset(dp, LL_INF, sizeof dp); dp[0][0] = 0, dp[0][1] = c[0]; bool ok = true; for(int i = 1; i < n; ++i){ if(s[i][0] >= s[i - 1][0]){ dp[i][0] = min(dp[i][0], dp[i - 1][0]); } if(s[i][1] >= s[i - 1][0]){ dp[i][1] = min(dp[i][1], dp[i - 1][0] + c[i]); } if(s[i][0] >= s[i - 1][1]){ dp[i][0] = min(dp[i][0], dp[i - 1][1]); } if(s[i][1] >= s[i - 1][1]){ dp[i][1] = min(dp[i][1], dp[i - 1][1] + c[i]); } if(dp[i][0] == LL_INF && dp[i][1] == LL_INF){ ok = false; break; } } if(ok){ printf("%I64d\n", min(dp[n - 1][0], dp[n - 1][1])); } else{ printf("-1\n"); } return 0; }
CodeForces - 706C Hard problem(dp+字符串)
标签:字典序 ble fine ack memset unsigned return tac ons
原文地址:http://www.cnblogs.com/tyty-Somnuspoppy/p/7121219.html