【后缀数组+字符串hash】 HDOJ 4029 Distinct Sub-matrix

先枚举宽度，然后把同一宽度的字符串用分隔符链接起来，然后求一下后缀数组，计算一下不同子串的个数。。。

#include <iostream>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 40005
#define eps 1e-10
#define mod 10000007
#define INF 2e9
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid  
#define rson o<<1 | 1, mid+1, R  
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
//head

const int x = 123;
char g[205][205];
int s[maxn];
ULL hash[205][205], xp[205];
map<ULL, int> mpp;
int sa[maxn], t1[maxn], t2[maxn], c[maxn];
int rank[maxn], height[maxn];
int n, m, tot, sn;

void build(int n, int m)
{
	int *x = t1, *y = t2, p;
	for(int i = 0; i < m; i++) c[i] = 0;
	for(int i = 0; i < n; i++) c[x[i] = s[i]]++;
	for(int i = 1; i < m; i++) c[i] += c[i-1];
	for(int i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;
	for(int k = 1; k <= n; k <<= 1) {
		p = 0;
		for(int i = n - k; i < n; i++) y[p++] = i;
		for(int i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
		for(int i = 0; i < m; i++) c[i] = 0;
		for(int i = 0; i < n; i++) c[x[y[i]]]++;
		for(int i = 1; i < m; i++) c[i] += c[i-1];
		for(int i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
		swap(x, y), x[sa[0]] = 0, p = 1;
		for(int i = 1; i < n; i++)
			x[sa[i]] = y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k] ? p-1 : p++;
		if(p >= n) break;
		m = p;
	}
}

void getheight(int n)
{
	int k = 0;
	for(int i = 0; i <= n; i++) rank[sa[i]] = i;
	for(int i = 0; i < n; i++) {
		if(k) k--;
		int j = sa[rank[i] - 1];
		while(s[i+k] == s[j+k]) k++;
		height[rank[i]] = k;
	}
}

void read(void)
{
	scanf("%d%d", &n, &m);
	getchar();
	for(int i = 0; i < n; i++) {
		for(int j = 0; j < m; j++)
			g[i][j] = getchar();
		getchar();
	}
}

ULL H(int now, int L, int kk)
{
	return hash[now][kk] - hash[now + L][kk] * xp[L];
}

void work(void)
{
	ULL t;
	int tt, ans = 0;
	xp[0] = 1;
	for(int i = 1; i <= n; i++) xp[i] = xp[i-1] * x;
	for(int i = 0; i < m; i++) hash[n][i] = 0;
	for(int i = 0; i < m; i++)
		for(int j = n-1; j >= 0; j--)
			hash[j][i] = hash[j+1][i] * x + g[j][i] - 'A';
	for(int i = 1; i <= n; i++) {
		mpp.clear();
		sn = 0, tot = 1;
		for(int j = 0; j + i <= n; j++) {
			for(int k = 0; k < m; k++) {
				t = H(j, i, k);
				if(!mpp.count(t)) mpp[t] = tt = tot++;
				else tt = mpp[t];
				s[sn++] = tt;
			}
			s[sn++] = tot++;
		}
		s[sn] = 0;
		build(sn+1, tot);
		getheight(sn);
		ans += m * (m+1) / 2 * (n - i + 1);
		for(int i = 1; i <= sn; i++) ans -= height[i];
	}
	printf("%d\n", ans);
}

int main(void)
{
	int _, __;
	while(scanf("%d", &_)!=EOF) {
		__ = 0;
		while(_--) {
			read();
			printf("Case #%d: ", ++__);
			work();
		}
	}
	return 0;
}