FZU

点击打开链接

Fat brother and Maze are playing a kind of special (hentai) game with two integers. All the digits of these two integers are in the range from 1 to 9. After they’ve got these two integers, they thought that these two numbers were not large enough! (You know that the young people always like the HUGE THING in order to have more pleasure) So they decide to use the digits of these two integers to make a new BIGGER integer. At the beginning of this game, the new integer has no digit, each time Fat Brother and Maze can choose one of the two initial integers (if this integer exists) and move its first digit to the end of the new integer. For instance, if the new integer is 233 and the two initial integers are 3154 and 1324 now, they can choose the first digit of the integer 3154, which is 3, and add it to the end of the new integer to make it become 2333. The two initial integers are 154 and 1324 now after this action. Also they can choose the first digit of the integer 1324 and add it to the end of the integer 233 and make it become 2331. This process goes until the two initial integers are all empty. Now Fat Brother and Maze would like to know the maximum number they could get after this special (hentai) game.

Input

The first line of the date is an integer T (1 <= T <= 102), which is the number of the text cases.

Then T cases follow, each case contains two integers N and M (1 <= N,M <= 100000) indicated the number of the digits of these two integers. Then a line with N digits indicates the first integer and a line with M digits indicates the second integer. Note that all the digits are in the range from 1 to 9.

In 90% of test cases, N, M <= 1000.

Output

For each case, output the case number first, and then output the integer Fat Brother and Maze would get, this integer should be as large as possible.

Sample Input

13 42 5 23 6 3 1

Sample Output

Case 1: 3632521

题意：给出两个数组，要求在不改变两个数组内部顺序的前提下合并两个数组，最后得到的数组的字典序最大。

思路：讲道理一开始只知道O(n^2)的解法，后来看题解才知道是用后缀数组，复杂度降低贼多啊！！！后缀数组的话感觉还是比较难理解的，具体的学习可以参照：后缀数组

然后题目就变得很明显了，具体的也是参照基神的模板写的后缀数组模板，感觉基神的代码真的很容易看懂，思路贼清晰的那种。。。

#include<stdio.h>#include<iostream>#include<set>#include<queue>#include<algorithm>#include<math.h>#include<string.h>#include<string>#include<vector>using namespace std;const int inf = 0x3f3f3f3f ;const int MX = 2e5 + 5;const int mod = 1e9 + 7;const int INF = 0x3f3f3f3f;int a[MX] , ans[MX];char s[MX];int SA[MX], R[MX], H[MX];int wa[MX], wb[MX], wv[MX], wc[MX];bool cmp(int *r, int a, int b, int l) {    return r[a] == r[b] && r[a + l] == r[b + l];}void Suffix(int *r, int n ,int m = 128) {    //int n = strlen(r) + 1;    int i, j, p, *x = wa, *y = wb, *t;    for(i = 0; i < m; i++) wc[i] = 0;    for(i = 0; i < n; i++) wc[x[i] = r[i]]++;    for(i = 1; i < m; i++) wc[i] += wc[i - 1];    for(i = n - 1; i >= 0; i--) SA[--wc[x[i]]] = i;    for(j = 1, p = 1; p < n; j *= 2, m = p) {        for(p = 0, i = n - j; i < n; i++) y[p++] = i;        for(i = 0; i < n; i++) if(SA[i] >= j) y[p++] = SA[i] - j;        for(i = 0; i < n; i++) wv[i] = x[y[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--) SA[--wc[wv[i]]] = y[i];        for(t = x, x = y, y = t, p = 1, x[SA[0]] = 0, i = 1; i < n; i++) {            x[SA[i]] = cmp(y, SA[i - 1], SA[i], j) ? p - 1 : p++;        }    }    int k = 0; n--;    for(i = 0; i <= n; i++) R[SA[i]] = i;    for(i = 0; i < n; i++) {        if(k) k--;        j = SA[R[i] - 1];        while(r[i + k] == r[j + k]) k++;        H[R[i]] = k;    }}//后缀数组的写法int main(){    int t , kk = 1;    scanf("%d" , &t) ;    while(t--){        int n , m ;        scanf("%d%d" , &n , &m) ;        for(int i = 0 ; i < n ; i ++)            scanf("%d" , &a[i]);        a[n] = 0;        for(int i = n + 1 ; i < m + n + 1 ; i ++)            scanf("%d" , &a[i]);        a[n+m+1] = 0 ;        int len = n + m + 1;        Suffix(a , n + m + 1);        int p1 = 0 , p2 = n + 1 , k = 0;        while(p1 < n && p2 < len){            if(a[p1] > a[p2])                ans[k++] = a[p1++] ;            else if(a[p1] < a[p2])                ans[k++] = a[p2++] ;            else if(R[p1] > R[p2])//相等的话比较rank就行                ans[k++] = a[p1++];            else                ans[k++] = a[p2++];        }        while(p1 < n)//最后如果还有多余的，再放进来            ans[k++] = a[p1++];        while(p2 < len)            ans[k++] = a[p2++];            printf("Case %d: " , kk++);        for(int i = 0 ; i < n + m ; i ++)            cout<<ans[i];        cout<<endl;    }    return 0;}