Codeforces Round #376(Div. 2) C dp

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题目传送门： http://codeforces.com/contest/706/problem/C

题意：先给n个数代表倒置字符串所耗能量，再给n个字符串，有先后顺序。字符串只能被倒置，不能被替换。输出使n个字符串成功按照字典序排列所需最低能量，如果无法实现输出-1

思路：这题就是一道简单的二维dp。第二维有两种状态，0代表不倒置，1代表倒置

状态转移方程：

if(str[i] >= str[i-1]) dp[i][0] = min(dp[i][0],dp[i-1][0]);if(str[i] >= str1[i-1]) dp[i][0] = min(dp[i][0],dp[i-1][1]);if(str1[i] >= str[i-1]) dp[i][1] = min(dp[i][1],dp[i-1][0] + c[i]);if(str1[i] >= str1[i-1]) dp[i][1] = min(dp[i][1],dp[i-1][1] + c[i]);

很简单的一道题做了半天。。还是我太vegetable了

代码如下：

#include <iostream>#include <algorithm>#include <cstring>#include <stdio.h>#include <string>#include <cmath>#include <queue>#include <set>#include <map>#include <stack>#include <bitset>using namespace std;#define   lson          l,m,rt<<1#define   rson          m+1,r,rt<<1|1#define   ll            long long#define   ull           unsigned long long#define   mem(n,v)      memset(n,v,sizeof(n))#define   MAX           100005#define   MAXN          200005#define   PI            3.1415926#define   E             2.718281828459#define   opnin         freopen("text.in.txt","r",stdin)#define   opnout        freopen("text.out.txt","w",stdout)#define   clsin         fclose(stdin)#define   clsout        fclose(stdout)#define   haha1         cout << "haha1"<< endl#define   haha2         cout << "haha2"<< endl#define   haha3         cout << "haha3"<< endlconst int    INF    =   0x3f3f3f3f;const ll     INFF   =   0x3f3f3f3f3f3f3f3f;const double pi     =   3.141592653589793;const double inf    =   1e18;const double eps    =   1e-8;const ll     mod    =   1e9+7;const ull    mx     =   133333331;/**************************************************************************/int n;ll c[MAX];string str[MAX];string str1[MAX];ll dp[MAX][2];int main(){    mem(c,0);    cin >> n;    for(int i=1;i<=n;i++){        dp[i][0] = dp[i][1] = INFF;    }    for(int i=1;i<=n;i++) scanf("%I64d",&c[i]);    for(int i=1;i<=n;i++){        cin >> str[i];        str1[i] = str[i];        reverse(str1[i].begin(),str1[i].end());    }    dp[1][0] = 0;    dp[1][1] = c[1];    for(int i=2;i<=n;i++){        if(str[i] >= str[i-1]) dp[i][0] = min(dp[i][0],dp[i-1][0]);        if(str[i] >= str1[i-1]) dp[i][0] = min(dp[i][0],dp[i-1][1]);        if(str1[i] >= str[i-1]) dp[i][1] = min(dp[i][1],dp[i-1][0] + c[i]);        if(str1[i] >= str1[i-1]) dp[i][1] = min(dp[i][1],dp[i-1][1] + c[i]);    ll Min = dp[n][0];    Min = min(Min,dp[n][1]);    if(Min == INFF) cout << "-1" <<endl;    else cout << Min << endl;}

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