Codeforces Round #367 (Div. 2)-C - Hard problem-DP

来源：互联网发布：kali linux安装输入法编辑：程序博客网时间：2024/06/02 06:48

题意：给字符串ｎ个，

能否通过逆序某些字符串，使得最后整体字典序递增。

每个字符串翻转需要花费，输出最小花费使得　整体字典序递增

dp[i][0]表示第ｉ个字符串选择不翻转的花费

dp[i]【１】表示要翻转的花费

简单dp一下就好了。。最后答案判断是否超过IN

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

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <map>#include <set>#include <vector>#include <iostream>using namespace std;const double pi=acos(-1.0);double eps=0.000001;typedef long long  ll;string ss[100123];string rev_ss[100123];ll c[100123];ll dp[100123][2];char tmp[100123];char tt[100123];int  ok[100123][2];int main(){    int n;    cin>>n;    for (int i=1; i<=n; i++) scanf("%lld",&c[i]);    for (int i=1; i<=n; i++)    {        scanf("%s",tmp);        ss[i]=tmp;        int len=strlen(tmp);        for (int j=0; j<len; j++)            tt[j]=tmp[len-j-1];        tt[len]=0;        rev_ss[i]=tt;    }    for (int i=1; i<=n; i++)        dp[i][0]=dp[i][1]=1e16;    dp[0][0]=dp[0][1]=0;    ok[0][1]        =ok[0][0]=1;    for (int i=1; i<=n; i++)    {        if (ok[i-1][0])        {            if (ss[i]>=ss[i-1])                dp[i][0]=min(dp[i][0],dp[i-1][0]),ok[i][0]=1;            if (rev_ss[i]>=ss[i-1])                dp[i][1]=min(dp[i][1],c[i]+dp[i-1][0]),ok[i][1]=1;        }        if (ok[i-1][1])        {            if (ss[i]>=rev_ss[i-1])                dp[i][0]=min(dp[i][0],dp[i-1][1]),ok[i][0]=1;            if (rev_ss[i]>=rev_ss[i-1])                dp[i][1]=min(dp[i][1],c[i]+dp[i-1][1]),ok[i][1]=1;        }    }    if (ok[n][0]==0&&            ok[n][1]==0)    {        printf("-1\n");        return 0;    }    printf("%lld\n",min(dp[n][0],dp[n][1]));    return 0;}

0 0