zoukankan      html  css  js  c++  java
  • 2019CCPC秦皇岛I题 Invoker(DP)

    Invoker

    Time Limit: 15000/12000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 117    Accepted Submission(s): 35


    Problem Description
    In dota2, there is a hero named Invoker. He has 3 basic skills in the game, which are Quas, Wex and Exort. Once he launches a basic skill, he will gain the corresponding element, where Quas gives "Q", Wex gives "W" and Exort gives "E".
    Invoker can't have more than 3 elements simultaneously. If he launches a basic skill when he already owns 3 elements, he will get the corresponding element and lose the element he gained the earliest.
    As can be seen, there are 10 unordered combinations of 3 elements in 3 types, each represents a special skill, which are as follows:

    • Cold Snap: unordered element combination "QQQ", denoted by "Y"

    • Ghost Walk: unordered element combination "QQW", denoted by "V"

    • Ice Wall: unordered element combination "QQE", denoted by "G"

    • EMP: unordered element combination "WWW", denoted by "C"

    • Tornado: unordered element combination "QWW", denoted by "X"

    • Alacrity: unordered element combination "WWE", denoted by "Z"

    • Sun Strike: unordered element combination "EEE", denoted by "T"

    • Forge Spirit: unordered element combination "QEE", denoted by "F"

    • Chaos Meteor: unordered element combination "WEE", denoted by "D"
    • Deafening Blast: unordered element combination "QWE", denoted by "B"

    When Invoker owns 3 elements, he can launch the invoking skill, denoted by "R", to gain the special skill according to the elements he currently owns. After invoking, the elements won't disappear, and the chronological order of the 3 elements won't change.
    Now given a sequence of special skills, you want to invoke them one by one with using the minimum number of basic skills(Q,W,E) and invoking skill(R). Print the minimum number in a single line.
    At the beginning, Invoker owns no elements. And you should re-invoke the special skills even if you have already invoked the same skills just now.
     
    Input
    Input a single line containing a string s (1 ≤ |s| ≤ 100 000) that only contains uppercase letters in {B, C, D, F, G, T, V, X, Y, Z}, denoting the sequence of special skills.
     
    Output
    Output a single line containing a positive integer, denoting the minimum number of skills to launch.
     
    Sample Input
    XDTBVV
     
    Sample Output
    15
    Hint
    One possible scheme is QWWREERERWQRQRR.
     
    Source
     
    参考代码:
    #include<bits/stdc++.h>
    using namespace std;
    const int size=1e5+5;
    const int inf=0x3f3f3f3f;
    char ss[size];
    int dp[size][5][5];
    int ans[size];
    char s[15][5]={"QQQ","QQW","QQE","WWW","QWW","WWE","EEE","QEE","WEE","QWE"};
    int stru[6][3]={{0,1,2},{0,2,1},{1,0,2},{1,2,0},{2,1,0},{2,0,1}};
    inline int id(char c)
    {
        if(c=='Q') return 0;
        if(c=='W') return 1;
        if(c=='E') return 2;
    }
    int ty[size];
    inline int swap(char c)
    {
        if(c=='Y') return 0;
        if(c=='V') return 1;
        if(c=='G') return 2;
        if(c=='C') return 3;
        if(c=='X') return 4;
        if(c=='Z') return 5;
        if(c=='T') return 6;
        if(c=='F') return 7;
        if(c=='D') return 8;
        if(c=='B') return 9;
    }
    int main()
    {
        while(~scanf("%s",ss+1))
        {
            int len=strlen(ss+1);
            for(int i=0;i<=len;i++)
            {
                for(int j=0;j<=2;j++)
                {
                    for(int k=0;k<=2;k++)
                    {
                        dp[i][j][k]=inf;
                    }
                }
            }
            for(int i=1;i<=len;i++)
            {
                ty[i]=swap(ss[i]);
                ans[i]=inf;
            }
            ans[1]=3;
            dp[1][id(s[ty[1]][0])][id(s[ty[1]][1])]=3;
            dp[1][id(s[ty[1]][0])][id(s[ty[1]][2])]=3;
            dp[1][id(s[ty[1]][1])][id(s[ty[1]][0])]=3;
            dp[1][id(s[ty[1]][1])][id(s[ty[1]][2])]=3;
            dp[1][id(s[ty[1]][2])][id(s[ty[1]][1])]=3;
            dp[1][id(s[ty[1]][2])][id(s[ty[1]][0])]=3;
            for(int i=2;i<=len;i++)
            {
                if(ty[i]==ty[i-1])
                {
                    for(int j=0;j<=2;j++)
                    {
                        for(int k=0;k<=2;k++)
                        {
                            dp[i][j][k]=dp[i-1][j][k];
                        }
                    }
                    ans[i]=ans[i-1];
                    continue;
                }
                dp[i][id(s[ty[i]][0])][id(s[ty[i]][1])]=3+ans[i-1];
                dp[i][id(s[ty[i]][0])][id(s[ty[i]][2])]=3+ans[i-1];
                dp[i][id(s[ty[i]][1])][id(s[ty[i]][0])]=3+ans[i-1];
                dp[i][id(s[ty[i]][1])][id(s[ty[i]][2])]=3+ans[i-1];
                dp[i][id(s[ty[i]][2])][id(s[ty[i]][1])]=3+ans[i-1];
                dp[i][id(s[ty[i]][2])][id(s[ty[i]][0])]=3+ans[i-1];
                for(int j=0;j<6;j++)
                {
                    dp[i][id(s[ty[i]][stru[j][1]])][id(s[ty[i]][stru[j][2]])]=min(dp[i][id(s[ty[i]][stru[j][1]])][id(s[ty[i]][stru[j][2]])],dp[i-1][id(s[ty[i]][stru[j][0]])][id(s[ty[i]][stru[j][1]])]+1);
                    for(int k=0;k<=2;k++)
                    dp[i][id(s[ty[i]][stru[j][1]])][id(s[ty[i]][stru[j][2]])]=min(dp[i][id(s[ty[i]][stru[j][1]])][id(s[ty[i]][stru[j][2]])],dp[i-1][k][id(s[ty[i]][stru[j][0]])]+2);
                    ans[i]=min(ans[i],dp[i][id(s[ty[i]][stru[j][1]])][id(s[ty[i]][stru[j][2]])]);
                }
            }
            printf("%d
    ",ans[len]+len);
        }
        return 0;
    }
    View Code
  • 相关阅读:
    eclipse中集成python开发环境
    取消eclipse英文单词拼写验证
    exe所在路径
    [转]关于Megatops BinCalc RPN计算器的说明
    WinDbg 蓝屏dump分析教程
    Delphi与Windows 7下的用户账户控制(UAC)机制 及 禁用兼容性助手
    【Delphi7】 解决“程序第一次可以正常编译,但再次编译的时候会报错,必须重新打开Delphi”的问题
    解决xftp远程连接中文乱码
    创建用资源管理器打开FTP位置
    收藏夹里的js
  • 原文地址:https://www.cnblogs.com/csushl/p/11604647.html
Copyright © 2011-2022 走看看