acdream 1401 Lempel-Ziv Compression

首先用sa+rmq预处理一下

比较巧妙的是一个贪心，一个点要么我们直接输出字符，要么就把它缩成尽可能长的串

然后从后忘前dp一下就出结果了

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;const int N = 5000;char s[N];int r[N], tx[N], ty[N], rs[N], ranks[N], sa[N], height[N], rmq[N][20]; //rs基数排序bool cmp(int *r, int a, int b, int len){return (r[a] == r[b]) && (r[a + len] == r[b + len]);}void suffix(int n, int m) //n为长度，最大值小于m{int i, j, p, *x = tx, *y = ty, *t;for(i = 0; i < m; ++i) rs[i] = 0;for(i = 0; i < n; ++i) { x[i] = r[i]; ++rs[x[i]];}for(i = 1; i < m; ++i) rs[i] += rs[i - 1];for(i = n - 1; i >= 0; --i) sa[--rs[x[i]]] = i;for(j = p = 1; p < n; j <<= 1, m = p) {for(p = 0, i = n - j; i < n; ++i) y[p++] = i;for(i = 0; i < n; ++i) { if(sa[i] >= j) y[p++] = sa[i] - j; }for(i = 0; i < m; ++i) rs[i] = 0;for(i = 0; i < n; ++i) ++rs[x[y[i]]];for(i = 1; i < m; ++i) rs[i] += rs[i - 1];for(i = n - 1; i >= 0; --i) sa[--rs[x[y[i]]]] = y[i];t = x, x = y, y = t;for(i = 1, p = 1, x[sa[0]] = 0; i < n; ++i) {if(cmp(y, sa[i - 1], sa[i], j)) x[sa[i]] = p - 1;else x[sa[i]] = p++;}}}void calheight(int n){int i, j, k = 0;for(i = 1; i <= n; ++i)ranks[sa[i]] = i;for(i = 0; i < n; ++i) {if(k)--k;j = sa[ranks[i] - 1];while(r[i + k] == r[j + k])++k;height[ranks[i]] = k;}}//////////////////////////////////////////////////////////////////////////rmq 求 lcpvoid initrmq(int n){int i, k;for(i = 2; i <= n; ++i)rmq[i][0] = height[i];for(k = 1; (1 << k) <= n; ++k) {for(i = 2; i + (1 << k) - 1 <= n; ++i) {rmq[i][k] = min(rmq[i][k - 1],rmq[i + (1 << (k - 1))][k - 1]);}}}int Log[N];void initlog(){Log[0] = -1;  for(int i=1;i<N;i++)  Log[i]=(i&(i-1))?Log[i-1]:Log[i-1] + 1;}int lcp(int a, int b ,int n)//求a,b的后缀的公用前缀长度，从0计{if(a==b) return n-a;a = ranks[a], b = ranks[b];if(a > b)swap(a, b);++a;int k = (int) Log[b - a + 1] / Log[2];return min(rmq[a][k], rmq[b - (1 << k) + 1][k]);}////////////////////////////////////////////////////////////////////////求第k小串（忽略重复）/*__int64 effectNum[N],sumEffectNum[N],allNum;//分别表示sa中的i后边有效串数，前i总有效串数;allNum为总共不同串的个数int aimSl,aimSr,aimSlength;//目标串左右端点以及长度void stStringInit(int n){sumEffectNum[0]=0;for(int i=1;i<=n;i++){effectNum[i]=n-sa[i]-height[i];sumEffectNum[i]=sumEffectNum[i-1]+effectNum[i];}allNum=sumEffectNum[n];}int aimSbs(int left,int right,__int64 v){int mid;while(left<right){mid=(left+right)>>1;if(sumEffectNum[mid]<v) left=mid+1;//if(不符合条件)else right=mid;}return left;}void getAimString(__int64 aimst,int n){int temp;temp=aimSbs(1,n,aimst);aimSl=sa[temp];aimSr=aimSl+height[temp]+aimst-sumEffectNum[temp-1]-1;aimSlength=aimSr-aimSl+1;}*///////////////////////////////////////////////////////////////////////////////////////////////int to[5000],chang[5000];bool vis[5000];int dp[5000];int main(){initlog();while( scanf("%s", s)!=EOF){int i,j,n;n = strlen(s);for(i = 0; i < n; ++i)r[i] = s[i];r[n] = 0;//便于比较suffix(n + 1, 128);calheight(n);initrmq(n);memset(chang,0,sizeof(chang));memset(to,-1,sizeof(to));for(i=0;i<n;i++)for(j=0;j<i;j++){if(lcp(i,j,n)>chang[i]){to[i]=j;chang[i]=lcp(i,j,n);}}memset(vis,0,sizeof(vis));memset(dp,0,sizeof(dp));for(i=n-1;i>=0;i--){if(to[i]==-1){vis[i]=1;dp[i]=dp[i+1]+9;}else{if( dp[i+1]+9 < dp[i+chang[i]]+25 ){vis[i]=1;dp[i]=dp[i+1]+9;}else{dp[i]=dp[i+chang[i]]+25;}}}printf("%d\n",dp[0]);for(i=0;i<n;){if(vis[i]==1){printf("%c",s[i]);i++;}else{printf("(%d,%d)",i-to[i],chang[i]);i+=chang[i];}}printf("\n");}return 0;}/*simple:ababcaaabcn=10st  string    ranks   /  sa   saString    height0  ababcaaabc   3     /  10   0            01  babcaaabc    6     /  5    aaabc        02  abcaaabc     5     /  6    aabc         23  bcaaabc      8     /  0    ababcaaabc   14  caaabc       10    /  7    abc          25  aaabc        1     /  2    abcaaabc     36  aabc         2     /  1    babcaaabc    07  abc          4     /  8    bc           18  bc           7     /  3    bcaaabc      29  c            9     /  9    c            010 0            0     /  4    caaabc       1*/