POJ 3274 Gold Balanced Lineup

Gold Balanced Lineup

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 7600 Accepted: 2245


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
Line 1: Two space-separated integers, N and K.
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
Line 1: A single integer giving the size of the largest contiguous balanced group of cows.

Sample Input
7 3
7
6
7
2
1
4
2

Sample Output
4

Hint
In the range from cow #3 to cow #6 (of size 4), each feature appears in exactly 2 cows in this range

Source
USACO 2007 March Gold

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAXE 100010
#define MAXN 100007

struct LEdge{
	int val[31], pos;
	int next;
}edge[MAXE];
int head[MAXN], nEdge;
int n, k;

void addEdge(int i, int* val, int pos){
	int j;
	for (j = 0; j < k; j++)
		edge[nEdge].val[j] = val[j];
	edge[nEdge].pos = pos;
	edge[nEdge].next = head[i];
	head[i] = nEdge++;
}

/*int hash(int *a){
	int i, v;
	for (i = v = 0; i < k; i++)
		v ^= a[i];
	return ((v % MAXN) + MAXN) % MAXN;
}

//折叠法
int hash(int *v){
	int i, s = 0;
	for(i = 0; i < k; i++)
		s=((s << 2) + (v[i] >> 4)) ^ (v[i] << 10);
	s = s % MAXN;
	s = s < 0 ? s + MAXN : s;
}
*/
/*
int ELFhash(char * key)
{
    unsigned int h = 0;
    while (*key){
        h = (h << 4) + *key++;
        unsigned int g = h & 0xf0000000L;
        if (g) h ^= g >> 24;
        h &= ~g;
    }
    return h % prime;
}

*/
int hash(int *a){
	int i, s = 0;
	for (i = 1; i < k; i++){
		s *= 41;
		s += a[i];
		s %= 100003;
		if (s < 0) s += 100003;
	}
	return s;
}

int equal(int *a, int *b){
	int i;
	for (i = 0; i < k; i++)
		if (a[i] != b[i]) break;
	if (i < k) return 0;
	return 1;
}

int find(int *a, int pos){
	int i, j, v;
	v = hash(a);
	j = pos;
	for (i = head[v]; i != -1; i = edge[i].next){
		if (equal(edge[i].val, a)){
			j = j > edge[i].pos ? edge[i].pos : j;
		}
	}
	addEdge(v, a, pos);
	return j;
}

int main(){
	int i, j, v, ans;
	int a[31], sum[31];
	while(scanf("%d%d", &n, &k) != EOF){
		memset(head, 0xff, sizeof(head));
		memset(sum, 0, sizeof(sum));
		nEdge = 0;
		ans = 0;
		find(sum, 0);
		for (i = 1; i <= n; i++){
			scanf("%d", &v);
			for (j = 0; j < k; j++){
				sum[j] += ((v & (1 << j)) ? 1 : 0);
			}
			for (j = 0; j < k; j++){
				a[j] = sum[j] - sum[0];
			}
			j = find(a, i);
			ans = ans < (i - j) ? (i - j) : ans;
		}
		printf("%d\n", ans);
	}
	return 0;
}
/*
1300MS...异或是慢啊还是分布不均啊...
靠换个人言NB的hash，竟然T掉了...奇怪了...
又换了个自称很简单的，竟然200MS+!
这个取数真好,100003就200MS+，100007就400MS+...
虽然模运算看起来很多= =，但分布均匀了

补充，这题得分析才能转换成匹配
要求s[i][l] - s[j][l]都相等, l = 0...k中i,j差最大的
即s[i][l] - s[j][l] = s[i][0] - s[j][0]
s[i][l] - s[i][0] = s[j][l] - s[j][0]
取a[i][l] = s[i][l] - s[i][0]
则求a[i][l] = a[j][l]中i,j差最大的，这就是匹配问题了
*/