codeforces 219C Color Stripe（貪心）

題目鏈接

C. Color Stripe

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A colored stripe is represented by a horizontal row of n square cells, each cell is pained one of k colors. Your task is to repaint the minimum number of cells so that no two neighbouring cells are of the same color. You can use any color from 1 to k to repaint the cells.

Input

The first input line contains two integers n and k (1 ≤ n ≤ 5·10⁵; 2 ≤ k ≤ 26). The second line contains n uppercase English letters. Letter "A" stands for the first color, letter "B" stands for the second color and so on. The first k English letters may be used. Each letter represents the color of the corresponding cell of the stripe.

Output

Print a single integer — the required minimum number of repaintings. In the second line print any possible variant of the repainted stripe.

Sample test(s)

input

6 3
ABBACC

output

2
ABCACA

input

3 2
BBB

output

1
BAB

題意：給一個字符串，只有K個字符，從‘A’開始。問最少修改多少個字符（字符只能是從A開始的K個），可以使字符串相鄰的字符都不相同。

題解：貪心。如果K大於2，直接修改偶數位上的字符即可。如果k==2，那麼只有兩種清楚ABAB....,BABAB.....，判斷那種情況更優即可。

代碼如下：（比賽時對於k==2的情況是用dp做的）

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<string.h>
#include<string>
#include<stdlib.h>
typedef __int64 LL;
typedef unsigned __int64 LLU;
const int nn=510000;
const int inf=0x3fffffff;
const LL mod=1000000007;
const LL inf64=(LL)inf*inf;
const double pi=acos(-1.0);
const double eps=1e-8;
using namespace std;
int n,k;
char s[nn];
int dp[nn][2];
int pre[nn][2];
void solve1()
{
    int i,ans=0;
    int ix;
    for(i=1;i<n;i++)
    {
        ix=s[i]-'A';
        if(s[i]==s[i-1])
        {
            ans++;
            ix=(s[i]-'A'+1)%k;
            if(i<n-1&&ix==s[i+1]-'A')
            {
                ix=(ix+1)%k;
            }
        }
        s[i]=ix+'A';
    }
    printf("%d\n",ans);
    printf("%s\n",s);
}
stack<int>sta;
void solve2()
{
    int i,j;
    for(i=0;i<k;i++)
    {
        if(s[0]-'A'==i)
        {
            dp[0][i]=0;
        }
        else
            dp[0][i]=1;
    }
    for(i=1;i<n;i++)
    {
        for(j=0;j<=k;j++)
        {
            dp[i][j]=dp[i-1][j^1]+((s[i]-'A')==j?0:1);
        }
    }
    if(dp[n-1][0]<dp[n-1][1])
    {
        printf("%d\n",dp[n-1][0]);
        int ix=n-1,fc=0;
        while(ix>=0)
        {
            sta.push(fc+'A');
            ix--;
            fc=fc^1;
        }
    }
    else
    {
        printf("%d\n",dp[n-1][1]);
        int ix=n-1,fc=1;
        while(ix>=0)
        {
            sta.push(fc+'A');
            ix--;
            fc=fc^1;
        }
    }
    while(sta.size())
    {
        printf("%c",sta.top());
        sta.pop();
    }
    puts("");
}
int main()
{
    int i;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        scanf("%s",s);
        if(k>2)
        {
            solve1();
        }
        else
            solve2();
    }
    return 0;
}