HDU

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Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 6170    Accepted Submission(s): 1902

Problem Description

Ladies and gentlemen, please sit up straight.
Don't tilt your head. I'm serious.

For n given strings S1,S2,⋯,Sn, labelled from 1 to n, you should find the largest i (1≤i≤n) such that there exists an integer j (1≤j<i) and Sj is not a substring of Si.

A substring of a string Si is another string that occurs in Si. For example, ``ruiz" is a substring of ``ruizhang", and ``rzhang" is not a substring of ``ruizhang".

 

Input

The first line contains an integer t (1≤t≤50) which is the number of test cases.
For each test case, the first line is the positive integer n (1≤n≤500) and in the following n lines list are the strings S1,S2,⋯,Sn.
All strings are given in lower-case letters and strings are no longer than 2000 letters.

 

Output

For each test case, output the largest label you get. If it does not exist, output −1.

 

Sample Input

45ababczabcabcdzabcd4youlovinyouaboutlovinyouallaboutlovinyou5dedefabcdabcdeabcdef3abaccc

 

Sample Output

Case #1: 4Case #2: -1Case #3: 4Case #4: 3

 

Source

2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

Recommend

wange2014

 

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题目大意:

给出n个字符串，找到最大的j使得从1->j-1中存在s[i]不是s[j]的子串。

思路：

如果直接枚举两个串复杂度为50*500*500=12500000,还不加匹配算法的复杂度,超时。

现在想可不可以剪枝。

两层循环，第一层i枚举所有子串,第二层j枚举母串。

如果s[i]是s[j]的子串，则s[i]串的贡献完全可以转化成s[j]串，则直接break到下一个子串.

如果s[i]串不是s[j]的子串，则s[j]串即为满足题目要求的，标记下来，以后遍历母串跳过s[j]即可。

然后遍历寻找最大的j即可。

KMP版代码:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 500 + 5;char s[maxn][2500];int t, n;int p[maxn], nxt[maxn][2500], len[maxn];void getNext(int t) {    int i, j;    j = nxt[t][0] = -1;    i = 0;    while(i < n) {        while(j != -1 && s[t][i] != s[t][j]) j = nxt[t][j];        nxt[t][++i] = ++j;    }}//判断s[v]是否包含s[u]int kmp(int v, int u) {    int i, j;    i = j = 0;    while(i < len[v]) {        while(j != -1 && s[v][i] != s[u][j]) j = nxt[u][j];        i++;j++;        if(j >= len[u]) {            return 1;        }    }    return 0;}int solve() {    memset(p,1,sizeof(p));    int ans = -1;    for(int i = 1; i < n; i++) {        for(int j = i + 1; j <= n; j++) {            if(p[j]) {                if(kmp(j, i)) break;                else {                    p[j] = 0;                    ans = max(ans, j);                }            }        }    }    return ans;}int main() {    scanf("%d", &t);    for(int kase = 0; kase < t; kase++) {        scanf("%d", &n);        for(int i = 1; i <= n; i++) {            scanf("%s", s[i]);            len[i] = strlen(s[i]);            getNext(i);        }        printf("Case #%d: %d\n",kase+1, solve());    }    return 0;}

看了别人的AC，才直到有strstr函数，可以寻找s[i]在s[j]串中的位置。

strstr版代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;using namespace std;const int maxn = 500 + 5;char s[maxn][2500];int t, n;int p[maxn];int solve() {    memset(p,1,sizeof(p));    int ans = -1;    for(int i = 1; i < n; i++) {        for(int j = i + 1; j <= n; j++) {            if(p[j]) {                if(strstr(s[j], s[i])) break;                else {                    p[j] = 0;                    ans = max(ans, j);                }            }        }    }    return ans;}int main() {    scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d", &n);        for(int i = 1; i <= n; i++)             scanf("%s", s[i]);        printf("Case #%d: %d\n",++kase, solve());    }    return 0;}