POJ3080Blue Jeans（暴力）
开始做字符串专题，地址
第一题水题，暴力就可以做
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype> 10 #include <cstring> 11 #include <cstdlib> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 #define INF 0x3f3f3f3f 16 #define inf (-((LL)1<<40)) 17 #define lson k<<1, L, mid 18 #define rson k<<1|1, mid+1, R 19 #define mem0(a) memset(a,0,sizeof(a)) 20 #define mem1(a) memset(a,-1,sizeof(a)) 21 #define mem(a, b) memset(a, b, sizeof(a)) 22 #define FOPENIN(IN) freopen(IN, "r", stdin) 23 #define FOPENOUT(OUT) freopen(OUT, "w", stdout) 24 25 template<class T> T CMP_MIN(T a, T b) { return a < b; } 26 template<class T> T CMP_MAX(T a, T b) { return a > b; } 27 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 28 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 29 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 30 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 31 32 //typedef __int64 LL; 33 typedef long long LL; 34 const int MAXN = 1000; 35 const int MAXM = 100005; 36 const double eps = 1e-12; 37 38 char str[20][100], ansStr[100]; 39 int T, N; 40 41 int maxLen(char* s1, char* s2) 42 { 43 int ans = 0; 44 for(int i=0;s2[i];i++) 45 { 46 int len = 0; 47 for(int j=0;s1[j];j++) 48 { 49 if(s2[i+j] == s1[j]) len++; 50 else break; 51 } 52 ans = max(ans, len); 53 } 54 return ans; 55 } 56 57 int main() 58 { 59 while(~scanf("%d", &T))while(T--) 60 { 61 mem0(str); mem0(ansStr); 62 scanf("%d%*c", &N); 63 for(int i=0;i<N;i++) scanf("%s", str[i]); 64 int ans = 0; 65 for(int i=0;str[0][i];i++) 66 { 67 int len = INF; 68 for(int j=1;j<N;j++) 69 { 70 len = min( len, maxLen(&str[0][i], str[j]) ); 71 } 72 if(ans < len) 73 { 74 ans = len; 75 for(int j=i;j<i+ans;j++) ansStr[j-i] = str[0][j]; 76 ansStr[i+ans] = 0; 77 } 78 else if(ans == len) 79 { 80 int ok = 0; 81 for(int j=0;j<ans;j++) 82 { 83 if(ok) ansStr[j] = str[0][i+j]; 84 else if(ansStr[j] != str[0][i+j]) 85 { 86 if(ansStr[j] < str[0][i+j]) break; 87 else { ansStr[j] = str[0][i+j]; ok = 1; } 88 } 89 } 90 } 91 } 92 if(ans >= 3) 93 printf("%s\n", ansStr ); 94 else printf("no significant commonalities\n"); 95 } 96 return 0; 97 }
