zoukankan      html  css  js  c++  java
  • HDU

    链接:

    https://vjudge.net/problem/HDU-2457

    题意:

    Biologists finally invent techniques of repairing DNA that contains segments causing kinds of inherited diseases. For the sake of simplicity, a DNA is represented as a string containing characters 'A', 'G' , 'C' and 'T'. The repairing techniques are simply to change some characters to eliminate all segments causing diseases. For example, we can repair a DNA "AAGCAG" to "AGGCAC" to eliminate the initial causing disease segments "AAG", "AGC" and "CAG" by changing two characters. Note that the repaired DNA can still contain only characters 'A', 'G', 'C' and 'T'.

    You are to help the biologists to repair a DNA by changing least number of characters.

    思路:

    DP,但是是在trie图上跑, 跟找没有模板穿串很像,构造出一个跟给的串差别最小的.
    DP[i][j], 表示长度i,以j节点结尾,不包含模板串的最小操作, 对某个节点,向下扩展一个判断在原串上是否需要修改, 选一个最小的.
    注意,fail节点指向的点的状态要向下更新, 忘记更新wa了好久..

    代码:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    //#include <memory.h>
    #include <queue>
    #include <set>
    #include <map>
    #include <algorithm>
    #include <math.h>
    #include <stack>
    #include <string>
    #include <assert.h>
    #include <iomanip>
    #include <iostream>
    #include <sstream>
    #define MINF 0x3f3f3f3f
    using namespace std;
    typedef long long LL;
    const LL MOD = 20090717;
    const LL MAXN = 2e6+10;
    const int MAXASCII = 4;
    
    struct TrieTree
    {
        int Next[MAXASCII];
        int end;
        int fail;
        void Clear()
        {
            memset(Next, 0, sizeof(Next));
            end = 0;
            fail = 0;
        }
    }tree[MAXN];
    
    char s[MAXN];
    int Dp[1010][1010];
    map<char, int> Mp;
    int n, m, k, cnt;
    
    void Insert(char *s)
    {
        int len = strlen(s);
        int p = 0;
        for (int i = 0;i < len;i++)
        {
            if (tree[p].Next[Mp[s[i]]] == 0)
                tree[p].Next[Mp[s[i]]] = ++cnt, tree[cnt].Clear();
            p = tree[p].Next[Mp[s[i]]];
        }
        tree[p].end = 1;
    }
    
    void BuildAC()
    {
        queue<int> que;
        for (int i = 0;i < MAXASCII;i++)
        {
            if (tree[0].Next[i] != 0)
            {
                tree[tree[0].Next[i]].fail = 0;
                que.push(tree[0].Next[i]);
            }
        }
        while (!que.empty())
        {
            int u = que.front();
            que.pop();
            if (tree[tree[u].fail].end)
                tree[u].end = 1;
            for (int i = 0;i < MAXASCII;i++)
            {
                if (tree[u].Next[i] != 0)
                {
                    tree[tree[u].Next[i]].fail = tree[tree[u].fail].Next[i];
                    que.push(tree[u].Next[i]);
                }
                else
                    tree[u].Next[i] = tree[tree[u].fail].Next[i];
            }
        }
    }
    
    int Solve()
    {
        int len = strlen(s+1);
        memset(Dp, MINF, sizeof(Dp));
        Dp[0][0] = 0;
        for (int i = 0;i < len;i++)
        {
            for (int j = 0;j < cnt;j++)
            {
                if (Dp[i][j] == MINF)
                    continue;
                for (int t = 0;t < 4;t++)
                {
                    if (tree[tree[j].Next[t]].end == 1)
                        continue;
                    int tmp = Dp[i][j] + (Mp[s[i+1]] == t?0:1);
                    Dp[i+1][tree[j].Next[t]] = min(Dp[i+1][tree[j].Next[t]], tmp);
                }
            }
        }
        int tmp = MINF;
        for (int i = 0;i <= cnt;i++)
            tmp = min(tmp, Dp[len][i]);
        return ((tmp == MINF)?-1:tmp);
    }
    
    int main()
    {
        Mp['A'] = 0;
        Mp['T'] = 1;
        Mp['G'] = 2;
        Mp['C'] = 3;
        int testcnt = 0;
        while (~scanf("%d", &n) && n)
        {
            cnt = 0;
            tree[cnt].Clear();
            for (int i = 1;i <= n;i++)
            {
                scanf("%s", s);
                Insert(s);
            }
            scanf("%s", s+1);
            BuildAC();
            printf("Case %d: %d
    ", ++testcnt, Solve());
        }
    
        return 0;
    }
    
  • 相关阅读:
    详细解说python垃圾回收机制
    Vue-- 监听路由参数变化,数据无法更新 解决方案
    解决“只能通过Chrome网上应用商店安装该程序”的方法 ---离线安装谷歌浏览器插件
    axios POST提交数据的三种请求方式写法
    axios POST提交数据的三种请求方式写法
    vue+element后台系统 自己动手撸(一)
    element-ui中 table表格hover 修改背景色
    解决vue的{__ob__: observer}取值问题
    Vue [__ob__: Observer]取不到值问题的解决
    VUE监听路由变化的几种方式
  • 原文地址:https://www.cnblogs.com/YDDDD/p/11647417.html
Copyright © 2011-2022 走看看