Codeforces GYM 100114 D. Selection 线段树维护DP

D. Selection

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100114

Description

When selecting files in an application dialog, Vasya noted that he can get the same selection in different ways. A simple mouse click selects a single file (the existing selection is discarded). A shift-click is used to select a range of files from the file clicked last time to the current file (the existing selection is discarded). Finally, a control-click is used to invert the selection state of a single file. Consider a sequence of actions. First we select file #5 simply by clicking it. Then, shift-clicking file #10 we get the following selection: #5, #6, #7, #8, #9, #10. If after that we control-click files #7 and #3 then we will have files #3, #5, #6, #8, #9, and #10 selected. Shift-clicking file #1 we select files #1, #2, and #3 (last time we clicked file #3, and the previous selection is gone). Vasya is wondering, what the minimum number of clicks will be, to make a certain selection from the list of files. Write a program to determine the optimal way of making the required selection. If there are several minimal solutions, any of them is considered correct. Example. Suppose we are to select files #2, #5, #6, #8, #9 from a list of 10 files. A possible optimal solution will include the following clicks: 5, Shift+9, Ctrl+2, Ctrl+7.

Input

The first line contains an integer n, the number of files in the list. The following line contains n characters defining the required selection. If i-th file is to be selected then there is an asterisk (“*”) in position i, and a dot (“.”) otherwise.

Output

The first line of the output file must contain a single integer k – the minimum number of clicks necessary to make the given selection. The following k lines must define the way to make such a selection. Each line should contain the number of file to be clicked on the corresponding step, and a prefix “Ctrl+” or “Shift+” (without quotation marks) where necessary.

Sample Input

10 .*..**.**.

Sample Output

4 5 Shift+9 Ctrl+2 Ctrl+7

HINT

1 ≤ n ≤ 105 .

题意

有3种操作

1.光标移动到X

2.shift X，直接选择从光标到X的位置

3.选择X，如果X已经被选中，那就取消X的选中状态

问你最少多少步，可以选择所有的*

并且把步骤输出

题解：

注意，只能shift 1次，所以直接扫一遍就好了

跑一遍线段树维护的DP，表示到这儿，所需要的最小代价是多少

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <iomanip>
#include <string>
#include <ctime>
#include <list>
#include <bitset>
typedef unsigned char byte;
#define pb push_back
#define input_fast std::ios::sync_with_stdio(false);std::cin.tie(0)
#define local freopen("in.txt","r",stdin)
#define pi acos(-1)

using namespace std;
const int maxn = 1e5 + ;
char str[maxn];
int length , sum[maxn], dp[maxn];
vector<int>Q,Q2;

struct operation
{
    int x;
    int type;
};

operation nxt[maxn];

struct QueryData
{
    int minv , minpos;
    QueryData(int minv , int minpos)
    {
        this->minv = minv , this->minpos = minpos;
    }
};

typedef int SgTreeDataType;
struct treenode
{
  int L , R  ;
  SgTreeDataType minv , minpos;
  void updata(SgTreeDataType v)
  {
        minv = v;
  }
};

treenode tree[maxn * ];

inline void push_up(int o)
{
    if(tree[o*].minv > tree[o*+].minv)
    {
        tree[o].minv = tree[o*+].minv;
        tree[o].minpos = tree[o*+].minpos;
    }
    else
    {
        tree[o].minv = tree[o*].minv;
        tree[o].minpos = tree[o*].minpos;
    }
}

inline void build_tree(int L , int R , int o)
{
    tree[o].L = L , tree[o].R = R,tree[o].minv = << , tree[o].minpos = ;
    if(L == R) tree[o].minpos = L;
    if (R > L)
    {
        int mid = (L+R) >> ;
        build_tree(L,mid,o*);
        build_tree(mid+,R,o*+);
    }
}

inline void updata(int QL,int QR,SgTreeDataType v,int o)
{
    int L = tree[o].L , R = tree[o].R;
    if (QL <= L && R <= QR) tree[o].updata(v);
    else
    {
        int mid = (L+R)>>;
        if (QL <= mid) updata(QL,QR,v,o*);
        if (QR >  mid) updata(QL,QR,v,o*+);
        push_up(o);
    }
}

inline QueryData query(int QL,int QR,int o)
{
    int L = tree[o].L , R = tree[o].R;
    if (QL <= L && R <= QR) return QueryData(tree[o].minv,tree[o].minpos);
    else
    {
        int mid = (L+R)>>;
        if (QL <= mid && QR > mid)
        {
            QueryData a = query(QL,QR,*o);
            QueryData b = query(QL,QR,*o+);
            if(a.minv < b.minv) return a;
            else return b;
        }
        else if (QL <= mid) return query(QL,QR,*o);
        else return query(QL,QR,*o+);
    }
}

void initiation()
{
    memset( dp ,  , sizeof(dp));
    scanf("%d%s",&length,str+);sum[] = ;
    for(int i =  ; i <= length ; ++ i)
    {
        sum[i] = sum[i-];
        if(str[i] == '*')
        {
            Q.push_back(i);
        }
        else
        {
            sum[i] ++ ;
            Q2.push_back(i);
        }
    }
}

void solve()
{
    int sz = Q.size();
    int ansL = Q[],ansR = Q[],ans=sz;
    build_tree(  , sz -  ,  );
    for(int i =  ; i < sz ; ++ i) updata( i , i , sum[Q[i]] - i , );
    for(int i =  ; i < sz - ; ++ i)
    {
        QueryData y = query( i +  , sz  - , );
        int newans = i +  + sz + y.minv - sum[Q[i]];
        if(newans < ans)
        {
            ans = newans;
            ansL = i;
            ansR = y.minpos;
        }
    }
    printf("%d\n",ans);
    if(ansL != ansR)
    {
        printf("%d\n",Q[ansL]);
        printf("Shift+%d\n",Q[ansR]);
        for(int i =  ; i < sz ; ++ i) if(i < ansL || i > ansR) printf("Ctrl+%d\n",Q[i]);
        for(int i =  ; i < Q2.size() ; ++ i) if( Q2[i] < Q[ansR] && Q2[i] > Q[ansL]) printf("Ctrl+%d\n",Q2[i]);
    }
    else
    {
        printf("%d\n",Q[]);
        for(int i =  ; i < sz ; ++ i) printf("Ctrl+%d\n",Q[i]);
    }
}

int main(int argc,char *argv[])
{
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    initiation();
    if(Q.size() == ) printf("0\n");
    else if(Q.size() == ) printf("1\n%d\n",Q[]);
    else if(Q.size() == ) printf("2\n%d\nCtrl+%d\n",Q[],Q[]);
    else solve();
    return ;
}