[USACO Training] Broken Necklace （DP）

Broken Necklace

题目大意：

有一串项链，有红色（r），白色（w），蓝色（b）组成，现在从任意位置把项链断开，从断开的两头分别向项链中间遍历。以左端为例，如果左端第一个为红色，那么从左开始取出所有红色，直到碰到蓝色停止。问最多可以从这串项链中取走多少珠子。（白色既可以当做红色，也可以当做蓝色）

解题思路：

O（N^2）

由于珠子数不多，最多350颗。那我们可以用纯暴力的方式暴搜，即从每一个可以拆开的地方断开项链，然后从两段遍历，得到一个值，并与当前最大值比较。

这里有一个技巧，把字符串复制一份连接到原先的后面，方便处理。

/*ID: [e-mail]: beadsLANG: C++*/// O(n^2)#include<iostream>#include<fstream>#include<cstdio>#include<cstring>#include<cmath>#include<map>#include<queue>#include<stack>#include<vector>#include<set>#include<ctype.h>#include<algorithm>#include<string>#define PI acos(-1.0)#define maxn 1000#define INF 1<<25#define mem(a, b) memset(a, b, sizeof(a))typedef long long ll;using namespace std;int main (){    string str;    int n, i, l, r, mx = 0;    ifstream fin ("beads.in");    ofstream fout ("beads.out");    fin>>n;    fin>>str;    str += str;    for (i = 0; i < n; i++)    {        l = r = 1;        int k = i + 1;        char flag;        while(l < n)        {            int j = i;            while(str[j] == 'w') j++;            flag = str[j];            if (str[k] == flag || str[k] == 'w') k++, l++;            else break;        }        k = i + n - 2;        while(r < n)        {            int j = i;            while(str[j] == 'w') j--;            flag = str[j];            if (str[k] == str[i + n - 1] || str[k] == 'w') k--, r++;            else break;        }        if (l + r > mx) mx = l + r;    }    if (mx > n) mx = n;    fout<<mx<<endl;    return 0;}

O（N）
但如果N比较大，这种做法的效率就很低下，我们可以用DP的思想来解决这道题，状态方程如下：（用r 表示红色，b 表示蓝色）而且从断开出从左遍历和从右遍历一样

r[0] = p[0] = 0 If c = 'r' then r[p+1] = r[p] + 1 and b[p+1] = 0        because the length of the blue beads is 0. if c = 'b' then b[p+1] = b[p] + 1 and r[p+1] = 0 if c = 'w' then both length of the red and length of blue beads             can be longer.so r[p+1] = r[p]+1 and b[p+1] = b[p] + 1.

/*ID: [e-mail]: beadsLANG: C++*/// O(N) DP#include<iostream>#include<fstream>#include<cstdio>#include<cstring>#include<cmath>#include<map>#include<queue>#include<stack>#include<vector>#include<set>#include<ctype.h>#include<algorithm>#include<string>#define PI acos(-1.0)#define maxn 1000#define INF 1<<25#define mem(a, b) memset(a, b, sizeof(a))typedef long long ll;using namespace std;int main (){    ifstream fin ("beads.in");    ofstream fout ("beads.out");    int i, j, n, mx = 0;    int left[810][2] = {0}, right[810][2] = {0};    string str;    fin>>n>>str;    str += str;    for (i = 1; i <= 2 * n; i++)    {        if (str[i - 1] == 'r')            left[i][0] = left[i - 1][0] + 1, left[i][1] = 0;        else if (str[i - 1] == 'b')            left[i][1] = left[i - 1][1] + 1, left[i][0] = 0;        else            left[i][1] = left[i - 1][1] + 1, left[i][0] = left[i - 1][0] + 1;    }    for (i = 2 * n - 1; i >= 0; i--)    {        if (str[i] == 'r')            right[i][0] = right[i + 1][0] + 1, right[i][1] = 0;        else if (str[i] == 'b')            right[i][1] = right[i + 1][1] + 1, right[i][0] = 0;        else            right[i][1] = right[i + 1][1] + 1, right[i][0] = right[i + 1][0] + 1;    }    for (i = 0; i < 2 * n; i++)        mx = max(mx, max(left[i][0], left[i][1]) + max(right[i][0], right[i][1]));    mx = min(mx, n);    fout<<mx<<endl;    return 0;}