UVa 12388 Anti-Rhyme Pairs

方法：hash 二分

最优化问题转化为判定性问题。先预处理每个string的前缀hash，然后对于每组query，二分答案。当两个prefix 的hash value相同的时候，稳妥的方法是用O（prefix.length())的方法检查这两个prefix是否相同（1.430s)； 本题采用 bkdr hash，如果当hash value相同就认定两个prefix 相同，也通过了testcase（0.290s)。

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);

const int maxn = 1e5;
string str[maxn];
vector<ull> hashstr[maxn];
int n, q;

bool valid(int a, int b, int mid)
{
    --a, --b, --mid;
    // to be safe, one should perform the substring test on the following line.
    return  (hashstr[a][mid] == hashstr[b][mid]);// && str[a].substr(0, mid+1)==str[b].substr(0,mid+1));
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;  cin >> T;
    repn(kase, T)
    {
        cin >> n;
        rep(i, n) cin >> str[i];
        rep(i, n)
        {
            hashstr[i].clear();
            ull cur = 0, len = str[i].length();
            rep(j, len)
            {
                cur = 131*cur + (ull)str[i][j];
                hashstr[i].pb(cur);
            }
        }
        cout << "Case " << kase << ":
";
        cin >> q;
        rep(i, q)
        {
            int x, y;   cin >> x >> y;
            int left = 1, right = min(str[x-1].length(), str[y-1].length()), mid, ans=0;
            while (left <= right)
            {
                mid = (left+right)>>1;
                if (valid(x,y,mid))
                {
                    ans = mid;
                    left = mid+1;
                }
                else
                    right = mid-1;
            }
            cout << ans << newline;
        }
    }
}
/*
 
 2
 5
 daffodilpacm daffodiliupc distancevector distancefinder distinctsubsequence 4
 1 2
 1 5
 3 4
 4 5
 2
 acm
 icpc
 2
 1 2
 2 2

*/