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poj 3274 -- Gold Balanced Lineup
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 12110 | Accepted: 3553 |
Description
Farmer John‘s N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For example, cows exhibiting feature #1 might have spots, cows exhibiting feature #2 might prefer C to Pascal, and so on.
FJ has even devised a concise way to describe each cow in terms of its "feature ID", a single K-bit integer whose binary representation tells us the set of features exhibited by the cow. As an example, suppose a cow has feature ID = 13. Since 13 written in binary is 1101, this means our cow exhibits features 1, 3, and 4 (reading right to left), but not feature 2. More generally, we find a 1 in the 2^(i-1) place if a cow exhibits feature i.
Always the sensitive fellow, FJ lined up cows 1..N in a long row and noticed that certain ranges of cows are somewhat "balanced" in terms of the features the exhibit. A contiguous range of cows i..j is balanced if each of the K possible features is exhibited by the same number of cows in the range. FJ is curious as to the size of the largest balanced range of cows. See if you can determine it.
Input
Lines 2..N+1: Line i+1 contains a single K-bit integer specifying the features present in cow i. The least-significant bit of this integer is 1 if the cow exhibits feature #1, and the most-significant bit is 1 if the cow exhibits feature #K.
Output
Sample Input
7 37672142
Sample Output
4
Hint
1 /*====================================================================== 2 * Author : kevin 3 * Filename : GlodBalancedLineup.cpp 4 * Creat time : 2014-07-24 11:39 5 * Description : 6 ========================================================================*/ 7 #include <iostream> 8 #include <algorithm> 9 #include <cstdio>10 #include <cstring>11 #include <queue>12 #include <cmath>13 #define clr(a,b) memset(a,b,sizeof(a))14 #define M 10000515 #define MM 1897916 using namespace std;17 vector<int>hash[M];18 int s[M][35],n,k;19 int ElfHash(char *name) //linux hash20 {21 int h = 0,g;22 while(*name){23 h = (h<<4) + *name++;24 g = h & 0xf0000000;25 if(g) h ^= g>>24;26 h &= ~g;27 }28 return (h%MM);29 }30 int slove(int ch[]) //转换成字符串并求hash值31 {32 char str[35];33 for(int i = 0; i < k; i++){34 str[i] = ‘F‘+ch[i];35 }36 str[k] = ‘\0‘;37 int key = ElfHash(str);38 return key;39 }40 bool judge(int l,int h) //判断相等41 {42 for(int i = 0; i < k; i++){43 if(s[l][i] != s[h][i]){44 return false;45 }46 }47 return true;48 }49 int Search(int key,int l) //查找50 {51 int len = hash[key].size(),max = 0,temp;52 for(int i = 0; i < len; i++){53 if(judge(l,hash[key][i])){54 temp = fabs(l-hash[key][i]);55 if(temp > max) max = temp;56 }57 }58 hash[key].push_back(l); //向vector里追加元素59 return max;60 }61 int main(int argc,char *argv[])62 {63 scanf("%d%d",&n,&k);64 int a = 0,max = 0;65 clr(s,0);66 for(int i = 1; i <= n; i++){67 scanf("%d",&a);68 for(int j = 0; j < k; j++){69 int t = a&1;70 if(t) s[i][j] = s[i-1][j]+1;71 else s[i][j] = s[i-1][j];72 a = a>>1;73 }74 }75 for(int i = 0; i <= n; i++){76 int q = s[i][0];77 for(int j = 0; j < k; j++){78 s[i][j] -= q;79 }80 int key = slove(s[i]);81 int temp = Search(key,i);82 if(temp > max) max = temp;83 }84 printf("%d\n",max);85 return 0;86 }
code2:用静态邻接表模拟拉链。跑了300ms左右
1 /*====================================================================== 2 * Author : kevin 3 * Filename : GlodBalancedLineup.cpp 4 * Creat time : 2014-07-24 11:39 5 * Description : 6 ========================================================================*/ 7 #include <iostream> 8 #include <algorithm> 9 #include <cstdio>10 #include <cstring>11 #include <queue>12 #include <cmath>13 #define clr(a,b) memset(a,b,sizeof(a))14 #define M 10000515 #define MM 1897916 using namespace std;17 struct EdgeNode18 {19 int to,next;20 };21 EdgeNode Edges[M];22 int s[M][35],n,k,head[M],cnt;23 void AddEdges(int i,int j,int cnt)24 {25 Edges[cnt].to = j;26 Edges[cnt].next = head[i];27 head[i] = cnt;28 }29 int ElfHash(char *name)30 {31 int h = 0,g;32 while(*name){33 h = (h<<4) + *name++;34 g = h & 0xf0000000;35 if(g) h ^= g>>24;36 h &= ~g;37 }38 return (h%MM);39 }40 int slove(int ch[])41 {42 char str[35];43 for(int i = 0; i < k; i++){44 str[i] = ‘F‘+ch[i];45 }46 str[k] = ‘\0‘;47 int key = ElfHash(str);48 return key;49 }50 bool judge(int l,int h)51 {52 for(int i = 0; i < k; i++){53 if(s[l][i] != s[h][i]){54 return false;55 }56 }57 return true;58 }59 int Search(int key,int l)60 {61 int Max = 0,temp;62 for(int k = head[key]; k != -1; k = Edges[k].next){63 if(judge(l,Edges[k].to)){64 temp = fabs(l-Edges[k].to);65 if(temp > Max) Max = temp;66 }67 }68 AddEdges(key,l,cnt++);69 return Max;70 }71 int main(int argc,char *argv[])72 {73 scanf("%d%d",&n,&k);74 int a = 0,max = 0;75 clr(s,0);76 clr(head,-1);77 cnt = 0;78 for(int i = 1; i <= n; i++){79 scanf("%d",&a);80 for(int j = 0; j < k; j++){81 int t = a&1;82 if(t) s[i][j] = s[i-1][j]+1;83 else s[i][j] = s[i-1][j];84 a = a>>1;85 }86 }87 for(int i = 0; i <= n; i++){88 int q = s[i][0];89 for(int j = 0; j < k; j++){90 s[i][j] -= q;91 }92 int key = slove(s[i]);93 int temp = Search(key,i);94 if(temp > max) max = temp;95 }96 printf("%d\n",max);97 return 0;98 }