后缀数组 - poj3294 Life Forms

题目：

http://poj.org/problem?id=3294

题意：

求n个字符串中，至少在一半以上的字符串中出现的最长公共子串

思路;

后缀数组处理的又一经典问题

在每个字符串末尾添加一个不同的，小于所有串的任一字符的字符作为分隔标记，然后把所有串拼起来，形成一个新的字符串，对这个字符串求后缀数组，然后二分枚举长度mid，将问题转化为判定问题是否存在长度为mid的，在一半以上的字符串中出现的公共子串，这个判定问题遍历一次height数组即可解决。将后缀序列视为若干组，每组应满足任意两个后缀的LCP大于等于mid，检查是否有一组中后缀属于的字符串个数大于n/2

代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

#define rep(i,n) for(int i = 0; i < n; i++)
const int MAXSIZE = 1e5+200;

int rk[MAXSIZE], sa[MAXSIZE], height[MAXSIZE], wa[MAXSIZE], res[MAXSIZE];
int w[MAXSIZE];
int len;
int n;

void getSa(int up) {
	int *k = rk, *id = height, *r = res, *cnt = wa;
	rep(i, up) cnt[i] = 0;
	rep(i, len) cnt[k[i] = w[i]]++;
	rep(i, up) cnt[i + 1] += cnt[i];
	for (int i = len - 1; i >= 0; i--) {
		sa[--cnt[k[i]]] = i;
	}
	int d = 1, p = 0;
	while (p < len){
		for (int i = len - d; i < len; i++) id[p++] = i;
		rep(i, len)  if (sa[i] >= d) id[p++] = sa[i] - d;
		rep(i, len) r[i] = k[id[i]];
		rep(i, up) cnt[i] = 0;
		rep(i, len) cnt[r[i]]++;
		rep(i, up) cnt[i + 1] += cnt[i];
		for (int i = len - 1; i >= 0; i--) {
			sa[--cnt[r[i]]] = id[i];
		}
		swap(k, r);
		p = 0;
		k[sa[0]] = p++;
		rep(i, len - 1) {
			if (sa[i] + d < len && sa[i + 1] + d < len && r[sa[i]] == r[sa[i + 1]] && r[sa[i] + d] == r[sa[i + 1] + d])
				k[sa[i + 1]] = p - 1;
			else k[sa[i + 1]] = p++;
		}
		if (p >= len) return;
		d <<= 1, up = p, p = 0;
	}
}

void getHeight() {
	int i, k, h = 0;
	rep(i, len) rk[sa[i]] = i;
	rep(i, len) {
		if (rk[i] == 0)
			h = 0;
		else {
			k = sa[rk[i] - 1];
			if (h) h--;
			while (w[i + h] == w[k + h]) h++;
		}
		height[rk[i]] = h;
	}
}

void getSuffix() {
	int up = 126;
	getSa(up + 1);
	getHeight();
}

bool vis[105];
int belong[MAXSIZE];

int solve(){
    int l = 0, r =1000;
    while (l<=r){
        int mid = (l+r)>>1;
        bool flag = false;
        memset(vis,0,sizeof(vis));
        int k=1; //计算每段中的字符串个数
        vis[sa[0]] = true;
        for (int i=1;i<len;++i){
            if (flag) break;
            if (height[i]>=mid){
                if (!vis[belong[sa[i]]]){
                    vis[belong[sa[i]]] = true;
                    k++;
                }
            }
            else{
                memset(vis,0,sizeof(vis));
                vis[belong[sa[i]]] = true;
                if (k>n/2) flag = true;
                k = 1;
            }
        }
        if (k>n/2) flag = true;
//        cout<<"mid: "<<mid<<" "<<flag<<endl;
        if (flag) l = mid+1;
            else r = mid-1;
    }
    return r;
}

void print(int ans){
    memset(vis,0,sizeof(vis));
    int k=1;
    vis[sa[0]] = true;
    for (int i=1;i<len;++i){
        if (height[i]>=ans){
            if (!vis[belong[sa[i]]]){
                vis[belong[sa[i]]] = true;
                k++;
            }
        }
        else{
            memset(vis,0,sizeof(vis));
            if (k>n/2){
                for (int j=sa[i-1];j<sa[i-1]+ans;++j){
                    putchar(w[j]-101+'a');
                }
                putchar(10);
            }
            vis[belong[sa[i]]] = true;
            k = 1;
        }
    }
    if (k>n/2){
        for (int j=sa[len-1];j<sa[len-1]+ans;++j){
            putchar(w[j]-101+'a');
        }
        putchar(10);
    }
}

char temp[1005];

int main(){
    int m;
    scanf("%d",&n);
    while (n!=0){
        len = 0;
        for (int i=1;i<=n;++i){
            scanf("%s",temp);
            for (char* j=temp;*j;++j){
                belong[len] = i;
                w[len++] = *j-'a'+101;

            }
            belong[len] = i;
            w[len++] = i;
        }


        getSuffix();
        int ans = solve();

        //cout<<"ans: "<<ans<<endl;

        if (ans) print(ans);
            else printf("?\n");
        scanf("%d",&n);
        putchar(10);
    }
    return 0;
}