UVA 12338 Anti-Rhyme Pairs（hash + 二分）题解

题意：给出两个字符串的最大相同前缀。

思路：hash是要hash，不hash是不可能的。hash完之后从头遍历判断超时然后陷入沉默，然后告诉我这能二分orz，二分完就过了，写二分条件写了半天。不要用数组储存hash值，会爆，开vector就行。

代码：

#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<cstring>
#include<string>
#include<sstream>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define ll long long
#define ull unsigned long long
using namespace std;
const int maxn = 100000+10;
const int seed = 131;
const int MOD = 100013;
const int INF = 0x3f3f3f3f;
vector<ull> ha[maxn];
int len[maxn];
char s[10005];
void HASH(int x){
    len[x] = strlen(s + 1);
    ha[x].clear();
    ha[x].push_back(0);
    for(int i = 1; i <= len[x]; i++){
        ha[x].push_back(ha[x][i - 1] * seed + s[i]);
    }
}
int main(){
    int T, Case = 1;
    scanf("%d", &T);
    while(T--){
        int n;
        scanf("%d", &n);
        for(int i = 1;i <= n;i++){
            scanf("%s", s + 1);
            HASH(i);
        }
        int q;
        printf("Case %d:\n", Case++);
        scanf("%d", &q);
        while(q--){
            int u, v;
            scanf("%d%d", &u, &v);
            int l = 1, r = min(len[u], len[v]), m, ans = 0;
            while(l <= r){
                m = (l + r) / 2;
                if(ha[u][m] == ha[v][m]){
                    l = m + 1;
                    ans = m;
                }
                else r = m - 1;
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}
/*
Sample Input
2
5
daffodilpacm
daffodiliupc
distancevector
distancefinder
distinctsubsequence
4
1 2
1 5
3 4
4 5
2
acm
icpc
2
1 2
2 2
Sample Output
Case 1:
8
1
8
4
Case 2:
0
4
*/