POJ 2176 Folding (字符串)

题意：个你一个字符串将其进行压缩，使得压缩后的串长度最短。

压缩规则：

• A sequence that contains a single character from 'A' to 'Z' is considered to be a folded sequence. Unfolding of this sequence produces the same sequence of a single character itself. 
  
• If S and Q are folded sequences, then SQ is also a folded sequence. If S unfolds to S' and Q unfolds to Q', then SQ unfolds to S'Q'. 
  
• If S is a folded sequence, then X(S) is also a folded sequence, where X is a decimal representation of an integer number greater than 1. If S unfolds to S', then X(S) unfolds to S' repeated X times.

例如：AAAAAAAAAABABABCCD 压缩为 10(A)2(BA)B2(C)D

题解：区间DP，类似于矩阵链乘。对每个区间进行处理的时候，相当于对该区间进行打包，记录下该区间的基本模式串，以及该基本模式串的重复次数，例如：AAAAA => 5(A)，基本模式串是"A"，重复次数为5。我们应该尽量取重复次数多的打包方式，这样可以最大限度的与左右区间进行合并从而减少字符串的总长度。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct Segment{    int l, r; //l是基本模式串的长度，r是重复次数    char s[101]; //基本模式串} dp[101][101];int diglen(int num) {    if(num < 10) return 1;    else if(num < 100) return 2;    else return 3;}Segment pack(const Segment &seg) //进行打包{    Segment ret;    int digNum = diglen(seg.r); //求数字的字符个数    if(seg.l * seg.r < seg.l + digNum + 2) //+2是因为括号有两个字符    {        ret.r = 1;        ret.l = seg.l * seg.r;        strcpy(ret.s, seg.s);        for(int i = 1; i < seg.r; i++)            strcat(ret.s, seg.s);    }    else    {        ret.r = 1;        ret.l = seg.l + digNum + 2;        sprintf(ret.s, "%d(%s)", seg.r, seg.s);    }    return ret;}int packlen(const Segment &seg) //返回打包后的最小长度{    return min(seg.l * seg.r, seg.l + 2 + diglen(seg.r));}int linklen(const Segment &seg1, const Segment &seg2, int &linkr) //返回将两个区间合并后的最小长度{    int len;    if(seg1.l == seg2.l && 0 == strcmp(seg1.s, seg2.s)) //基本模式串一样则只需将重复次数相加    {        linkr = seg1.r + seg2.r;        len = seg1.l;    }    else     {        linkr = 1;        len = packlen(seg1) + packlen(seg2);    }    return min(linkr * len, len + 2 + diglen(linkr));}Segment link(const Segment &seg1, const Segment &seg2) //将两个区间的串进行合并{    Segment ret;    if(seg1.l == seg2.l && 0 == strcmp(seg1.s, seg2.s))    {        strcpy(ret.s, seg1.s);        ret.l = seg1.l;        ret.r = seg1.r + seg2.r;    }    else //基本模式串不同则需要先分别对两个区间进行打包，然后连接起来    {        Segment tmp1 = pack(seg1);         Segment tmp2 = pack(seg2);        strcpy(ret.s, tmp1.s);        strcat(ret.s, tmp2.s);        ret.l = tmp1.l + tmp2.l;        ret.r = 1;    }    return ret;}int main(){    char str[110];    while(scanf("%s",str) != EOF)    {        int i, j, k;        int len = strlen(str);        for(i = 0; i < len; i++)        {            dp[i][i].l = dp[i][i].r = 1;            dp[i][i].s[0] = str[i];            dp[i][i].s[1] = '\0';        }        int tmpl, tmpr;        int markl, markr, markj;        for(k = 1; k < len; k++)        {            for(i = 0; i + k < len; i++)            {                markj = i;                markl = linklen(dp[i][i], dp[i+1][i+k], markr);                for(j = i + 1; j < i + k; j++)                {                    tmpl = linklen(dp[i][j], dp[j+1][i+k], tmpr);                    if(tmpl < markl || (tmpl == markl && tmpr > markr))                        markl = tmpl, markr = tmpr, markj = j;                }                dp[i][i+k] = link(dp[i][markj], dp[markj+1][i+k]);            }        }        Segment tmp = pack(dp[0][len-1]);        printf("%s\n",tmp.s);    }    return 0;}