【HDOJ】4029 Distinct Sub-matrix

思路是枚举矩阵列数，然后将字符矩阵转换成字符串，通过字符数组求不同子串数目。最后，减去不成立的情况。使用特殊字符分割可能的组合。

/* 4029 */
#include <iostream>
#include <sstream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000")

#define sti                set<int>
#define stpii            set<pair<int, int> >
#define mpii            map<int,int>
#define vi                vector<int>
#define pii                pair<int,int>
#define vpii            vector<pair<int,int> >
#define rep(i, a, n)     for (int i=a;i<n;++i)
#define per(i, a, n)     for (int i=n-1;i>=a;--i)
#define clr                clear
#define pb                 push_back
#define mp                 make_pair
#define fir                first
#define sec                second
#define all(x)             (x).begin(),(x).end()
#define SZ(x)             ((int)(x).size())
#define lson            l, mid, rt<<1
#define rson            mid+1, r, rt<<1|1

const int maxr = 130;
const int maxn = 2e5+5;
const int base = 10007;
map<int,int> tb;
int a[maxn*3];
int height[maxn], rrank[maxn], sa[maxn*3];
int wa[maxn], wb[maxn], wc[maxn], wv[maxn];
char s[maxr][maxr];
int H[maxr][maxr];
int r, c;

bool c0(int *r, int a, int b) {
    return r[a]==r[b] && r[a+1]==r[b+1] && r[a+2]==r[b+2];
}

bool c12(int k, int *r, int a, int b) {
    if (k == 2)
        return r[a]<r[b] || (r[a]==r[b] && c12(1, r, a+1, b+1));
    else
        return r[a]<r[b] || (r[a]==r[b] && wv[a+1]<wv[b+1]);
}

void sort(int *r, int *a, int *b, int n, int m) {
    int i;
    
    for (i=0; i<n; ++i) wv[i] = r[a[i]];
    for (i=0; i<m; ++i) wc[i] = 0;
    for (i=0; i<n; ++i) wc[wv[i]]++;
    for (i=1; i<m; ++i) wc[i] += wc[i-1];
    for (i=n-1; i>=0; --i) b[--wc[wv[i]]] = a[i];
}

#define F(x) ((x)/3 + ((x)%3==1 ? 0:tb))
#define G(x) ((x)<tb ? (x)*3+1 : ((x)-tb)*3+2)
void dc3(int *r, int *sa, int n, int m) {
    int i, j, *rn=r+n, *san=sa+n, ta=0, tb=(n+1)/3, tbc=0, p;
    
    r[n] = r[n+1] = 0;
    for (i=0; i<n; ++i) if (i%3!=0) wa[tbc++] = i;
    sort(r+2, wa, wb, tbc, m);
    sort(r+1, wb, wa, tbc, m);
    sort(r, wa, wb, tbc, m);
    for (p=1, rn[F(wb[0])]=0, i=1; i<tbc; ++i)
        rn[F(wb[i])] = c0(r, wb[i-1], wb[i]) ? p-1 : p++;
    if (p < tbc)
        dc3(rn, san, tbc, p);
    else
        for (i=0; i<tbc; ++i) san[rn[i]] = i;
    for (i=0; i<tbc; ++i)
        if (san[i] < tb)
            wb[ta++] = san[i] * 3;
    if (n%3 == 1)
        wb[ta++] = n - 1;
    sort(r, wb, wa, ta, m);
    for (i=0; i<tbc; ++i) wv[wb[i]=G(san[i])] = i;
    for (i=0,j=0,p=0; i<ta && j<tbc; ++p)
        sa[p] = c12(wb[j]%3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
    while (i < ta) sa[p++] = wa[i++];
    while (j < tbc) sa[p++] = wb[j++];
}

void calheight(int *r, int *sa, int n) {
    int i, j, k = 0;
    
    for (i=1; i<=n; ++i) rrank[sa[i]] = i;
    for (i=0; i<n; height[rrank[i++]]=k)
    for (k?k--:0, j=sa[rrank[i]-1]; r[j+k]==r[i+k]; ++k) ;
}

void printSa(int n) {
    for (int i=1; i<=n; ++i)
        printf("%d ", sa[i]);
    putchar('
');
}

void printHeight(int n) {
    for (int i=1; i<=n; ++i)
        printf("%d ", height[i]);
    putchar('
');
}

void solve() {
    __int64 ans = 0, tot, tmp;
    int n = 0;
    
    memset(H, 0, sizeof(H));
    rep(w, 1, c+1) {
        tb.clr();
        int cn = c - w + 1;
        int cnt = cn + 1;
        int sidx = 1;
        
        rep(i, 0, r) {
            rep(j, 0, cn) {
                H[i][j] = H[i][j] * base + s[i][j+w-1]-'A';
                if (tb.find(H[i][j]) == tb.end())
                    tb[H[i][j]] = cnt++;
            }
        }
        
        n = 0;
        rep(j, 0, cn) {
            rep(i, 0, r)
                a[n++] = tb[H[i][j]];
            a[n++] = sidx++;
        }
        a[n] = 0;
        
        dc3(a, sa, n+1, cnt+5);
        calheight(a, sa, n);
        
        tot = 0;
        rep(i, 1, n+1)
            tot += (n - sa[i] - height[i]);
        
        tmp = n;
        rep(i, 0, cn) {
            tmp -= r;
            tot -= (r+1) * tmp;
            --tmp;
        }
        
        ans += tot;
    }
    
    printf("%I64d
", ans);
}

int main() {
    ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
        freopen("data.in", "r", stdin);
        freopen("data.out", "w", stdout);
    #endif
    
    int t;
    
    scanf("%d", &t);
    rep(tt, 1, t+1) {
        scanf("%d %d", &r, &c);
        rep(i, 0, r)
            scanf("%s", s[i]);
        printf("Case #%d: ", tt);
        solve();
    }
    
    #ifndef ONLINE_JUDGE
        printf("time = %d.
", (int)clock());
    #endif
    
    return 0;
}

数据生成器。

from random import randint, shuffle
import shutil
import string


def GenDataIn():
    with open("data.in", "w") as fout:
        t = 20
        uc = list(string.uppercase)
        fout.write("%d
" % (t))
        for tt in xrange(t):
            n = randint(20, 50)
            m = randint(20, 50)
            fout.write("%d %d
" % (n, m))
            for i in xrange(n):
                line = ""
                for j in xrange(m):
                    idx = randint(0, 25)
                    line += uc[idx]
                fout.write("%s
" % (line))
                
                
def MovDataIn():
    desFileName = "F:eclipse_prjworkspacehdojdata.in"
    shutil.copyfile("data.in", desFileName)

    
if __name__ == "__main__":
    GenDataIn()
    MovDataIn()