HDU 4691 The Unsolvable Problem 后缀数组

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
; ++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
= 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;
}