HDU 5172 GTY's gay friends 线段树 or Hash

统计一个区间中没有重复得数,用线段树搞法大概就是记录一下每个数上一次出现的位置,然后肯这个区间的pre的最大值是否大于左端点的值了。

//预处理前缀和+线段树解法
#include <cstdio>
#include <cstring>
#include <climits>
#include <cmath>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>

using namespace std;

typedef long long LL;

#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r

const int maxn = 1e6 + 10;
int n, q, arr[maxn], pre[maxn], maxv[maxn << 2], pos[maxn];
LL sum[maxn];

void build(int rt, int l, int r) {
	int mid = l + r >> 1;
	if(l == r) maxv[rt] = pre[l];
	else {
		build(lson); build(rson);
		maxv[rt] = max(maxv[rt << 1], maxv[rt << 1 | 1]);
	}
}

int ask(int rt, int l, int r, int ql, int qr) {
	if(ql <= l && qr >= r) return maxv[rt];
	int ret = -1, mid = l + r >> 1;
	if(ql <= mid) ret = max(ret, ask(lson, ql, qr));
	if(qr > mid) ret = max(ret, ask(rson, ql, qr));
	return ret;
}

void preprocess() {
	memset(pos, -1, sizeof(pos));
	for(int i = 1; i <= n; i++) {
		sum[i] = sum[i - 1] + arr[i];
		pre[i] = pos[arr[i]];
		pos[arr[i]] = i;
	}
	build(1, 1, n);
}

void gao(int l, int r) {
	LL len = r - l + 1, tsum = (1 + len) * len / 2;
	int tp = ask(1, 1, n, l, r);
	if(tsum == sum[r] - sum[l - 1] && tp < l) puts("YES");
	else puts("NO");
}

int main() {
	while(scanf("%d%d", &n, &q) != EOF) {
		for(int i = 1; i <= n; i++) scanf("%d", &arr[i]);
		preprocess();
		int l, r;
		for(int i = 1; i <= q; i++) {
			scanf("%d%d", &l, &r);
			gao(l, r);
		}
	}
	return 0;
}

 这里主要用到的hash姿势就是把每个数字映射到另外一个随机数上面,然后异或来搞就好了,不过要注意^运算的优先级.保险起见我判了两次。

#include <cstdio>
#include <cstring>
#include <climits>
#include <cmath>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <ctime>


typedef unsigned long long LL;
const int maxn = 1e6 + 10;
int n, q, arr[maxn];
LL sum[maxn], hash[maxn], nmap[maxn], thash[maxn], thash1[maxn], nmap1[maxn], hash1[maxn];

LL longRand() {
    return
        ((LL)rand() << 49) |
        ((LL)rand() << 34) |
        ((LL)rand() << 19) |
        ((LL)rand() << 4)  |
        (rand() & 15);
}

void init() {
    for(int i = 1; i <= 1000000; i++) {
        nmap[i] = longRand();
        nmap1[i] = longRand();
        thash[i] = thash[i - 1] ^ nmap[i];
        thash1[i] = thash1[i - 1] ^ nmap1[i];
    }
}

int main() {
    srand(time(NULL));
    init();
    while(scanf("%d%d", &n, &q) != EOF)  {
        for(int i = 1; i <= n; i++) {
            scanf("%d", &arr[i]);
            sum[i] = sum[i - 1] + arr[i];
            hash[i] = hash[i - 1] ^ nmap[arr[i]];
            hash1[i] = hash1[i - 1] ^ nmap1[arr[i]];
        }
        int l, r;
        for(int i = 1; i <= q; i++) {
            scanf("%d%d", &l, &r);
            LL len = r - l + 1;
            if(sum[r] - sum[l - 1] == len * (len + 1) / 2
                    && (hash[r] ^ hash[l - 1]) == thash[len]
                    && (hash1[r] ^ hash1[l - 1]) == thash1[len]
                    ) {
                puts("YES");
            }
            else puts("NO");
        }
    }
    return 0;
}

  

posted @ 2015-02-08 21:06  acm_roll  阅读(238)  评论(0编辑  收藏  举报