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CodeForces - 706C Hard problem(dp+字符串)
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)
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