POJ 3294 Life Forms 二分 + 哈希
终于搞死这题了,先是TLE,后是MLE, 然后是WA,AC得的不容易啊。。多项式差值取模神奇的算法。。。1280ms
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<iostream> #include<algorithm> #include<string> #include<vector> #include<set> using namespace std; #define Prime 113 #define MOD 5210 int N, yes; char DNA[112][1210]; int hash[1020]; //统计长度 struct Anode { unsigned long long key; //HASH值 int Num; //所树字符串编号 int cnt; //出现次数 }info; vector<Anode>Phash[MOD]; set<string>Myset, My; unsigned long long Fly[1010]; //unsigned long long my[1010]; /* 算法: 1.二分枚举长度K( 1, len ) you are to find the longest substring that is shared by more than half of them. 2. HASH表,给所有DNA字符串,建立一个HASH表,长度为K, 用Vector[i]存储第I个DNA字符串的所有长度为K的字串长度 3. 枚举 每个DNA长度为K的字符串 4.若一半以上有,就l = mid + 1, 继续枚举 */ void init( ) { for( int i = 0; i < MOD; i++) Phash[i].clear(); yes = 0; } //key, 字符串号, 当前长度 int get_hash(int Num, unsigned long long key,int k, char *str) { int x = key % MOD; int n = (int) Phash[x].size( ); int flag = 0; for(int i = 0; i < n; i++) { info = Phash[x][i]; if( info.key == key ) { flag = 1; if( info.Num != Num ) { Phash[x][i].cnt++; Phash[x][i].Num = Num; if( Phash[x][i].cnt > N / 2 ) { yes = 1; My.insert(str); //printf("k =%d here:%s\n",k,str); return 1; } } } } if( !flag ) { info.key = key; info.Num = Num; info.cnt = 1; Phash[x].push_back(info); } return 0; } void get_sub( int t, int x ) { int length = strlen(DNA[t]); char temp[1010]; char save[1010]; unsigned long long ans = 0; if( length < x ) return; for( int i = 0; i < x; i++) { ans = ans + (DNA[t][i] - 'a' + 1) * Fly[x-i-1]; temp[i] = DNA[t][i]; } temp[x] = '\0'; strcpy(save, temp); get_hash( t, ans, x , temp); for( int i = 1; i < length; i++) { if( i + x > length) break; ans -= (DNA[t][i-1] - 'a' + 1) * Fly[x-1]; ans *= Prime; ans += DNA[t][x + i - 1] - 'a' + 1; save[x+i-1] = DNA[t][x+i-1]; int u = 0; for( int j = i; j <= x + i - 1; j++) { temp[u++] = save[j]; } temp[u] = '\0'; get_hash( t, ans, x ,temp); } } int solve( int x ) { init( ); My.clear( ); for( int i = 1; i <= N; i++) get_sub(i, x); if( yes ) { Myset.clear(); Myset = My; return 1; } return 0; } int find( int l, int r ) { int ans = 0, vt = 0; while( l <= r ) { int mid = (l + r) / 2; if ( solve(mid) ) { ans = mid; l = mid + 1; } else r = mid - 1; } return ans; } int main( ) { int vv = 0; Fly[0] = 1; for( int i = 1; i <= 1000; i++) Fly[i] = Fly[i-1] * Prime; while( scanf("%d",&N),N) { int maxn = 0; memset(hash, 0, sizeof(hash)); for( int i = 1; i <= N; i++) { scanf("%s", DNA[i]); int len = strlen(DNA[i]); hash[len]++; if( len > maxn ) maxn = len; } int f = 0; //找到有效区间范围 for( int x = maxn; x >= 1; x--) { if ( hash[x] > N/2 ) { f = x; break; } else hash[x-1] += hash[x]; } int ans = find(1, maxn); if( vv ) puts(" "); vv = 1; if( ans == 0 ) puts("?"); else { set<string>::iterator its; for( its = Myset.begin(); its != Myset.end(); its++) cout<<*its<<endl; } } return 0; }
posted on 2012-07-22 07:29 more think, more gains 阅读(149) 评论(0) 编辑 收藏 举报