HDU 4691 The Unsolvable Problem 后缀数组

来源：互联网发布：网民网络素养编辑：程序博客网时间：2024/05/16 05:58

题意：给出一段文章，按照题目所给的方式进行压缩。求出压缩前和压缩后文章的长度。

思路：看到从任意位置匹配，想到了后缀数组。而查找最长公共前缀正式后缀数组干的事。

剩下的按照题目中进行计算就行了。

还是算比较裸的一道题。

代码如下：

#include <cstring>#include <cstdio>#include <algorithm>using namespace std;struct suffix_array{    static const int maxn =1000100;    int sa[maxn], rank[maxn], height[maxn];    int wa[maxn], wb[maxn], wv[maxn], wd[maxn];    int len;    int cmp(int *r, int a, int b, int l){        return r[a] == r[b] && r[a+l] == r[b+l];    }    void da(int *r, int n, int m){          //  倍增算法 r为待匹配数组  n为总长度 m为字符范围        int i, j, p, *x = wa, *y = wb, *t;        for(i = 0; i < m; i ++) wd[i] = 0;        for(i = 0; i < n; i ++) wd[x[i]=r[i]] ++;        for(i = 1; i < m; i ++) wd[i] += wd[i-1];        for(i = n-1; i >= 0; i --) sa[-- wd[x[i]]] = i;        for(j = 1, p = 1; p < n; j *= 2, 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 < n; i ++) wv[i] = x[y[i]];            for(i = 0; i < m; i ++) wd[i] = 0;            for(i = 0; i < n; i ++) wd[wv[i]] ++;            for(i = 1; i < m; i ++) wd[i] += wd[i-1];            for(i = n-1; i >= 0; i --) sa[-- wd[wv[i]]] = y[i];            for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i ++){                x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p - 1: p ++;            }        }    }    void calc_height(int *r, int n){           //  求height数组。        int i, j, k = 0;        for(i = 1; i <= n; i ++) rank[sa[i]] = i;        for(i = 0; i < n; height[rank[i ++]] = k){            for(k ? k -- : 0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k ++);        }    }/**************************************/    static const int MAX = 200100;    int p[MAX];    int d[MAX][20];    void rmq_init(int n){        p[0] = -1;        for(int i = 1; i <= n; ++i)            p[i] = i & (i-1)?p[i-1]:p[i-1]+1;        for(int i = 1; i <= n; ++i) d[i][0] = height[i];        for(int j = 1; j <= p[n]; ++j)            for(int i = 1; i + (1 << j) - 1 <= n; ++i)                d[i][j] = min(d[i][j-1],d[i+(1<<j-1)][j-1]);    }    int rmp_query(int l, int r){        int k = p[r - l + 1];        return min(d[l][k],d[r - (1<<k) + 1][k]);    }    int lcp(int l, int r){        if(l == r) return len - l + 1;        l = rank[l], r = rank[r];        if(l > r) swap(l,r);l++;        return rmp_query(l,r);    }/************************************************/    void calc(int*r, int n, int m){        len = n;        r[n] = 0;        da(r,n+1,m);        calc_height(r,n);        rmq_init(n);    }    int tolen(int n)    {        int len = 0;        while(n){            len++;            n /= 10;        }        if(len == 0) len++;        return len;    }    void solve(){        int T,a,b;        int prea = -1, prel = 0,nowl;        long long ans1 = 0, ans2 = 0;        scanf("%d",&T);        while(T--){            scanf("%d%d",&a,&b);            nowl = b - a;            ans1 += nowl + 1;            if(prea == -1)                ans2 += nowl + 3;            else{                long long com = min(lcp(prea,a),min(prel,nowl));                ans2 += tolen(com) + 2 + (nowl - com);                //printf("%d\n",com);            }            prea = a, prel = nowl;        }        printf("%I64d %I64d\n",ans1,ans2);    }} solver;char str[100100];int r[100100];int main(void){    //freopen("input.txt","r",stdin);    while(scanf("%s",str) != EOF){        int len = strlen(str);        copy(str,str+len,r);        solver.calc(r,len,256);        solver.solve();    }    return 0;}

0 0