填坑-回溯-预习 之 二分-尺取大总结

时间:2022-07-24
本文章向大家介绍填坑-回溯-预习 之 二分-尺取大总结,主要内容包括其使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。

尺取法

尺取法通常是对数组保存一对下标,即所选取的区间的左右端点,然后根据实际情况不断地推进区间左右端点以得出答案。尺取法比直接暴力枚举区间效率高很多,尤其是数据量大的时候,所以说尺取法是一种高效的枚举区间的方法,是一种技巧,一般用于求取有一定限制的区间个数或最短的区间等等。当然任何技巧都存在其不足的地方,有些情况下尺取法不可行,无法得出正确答案,所以要先判断是否可以使用尺取法再进行计算。

使用尺取法时应清楚以下四点:

1、 什么情况下能使用尺取法? 2、何时推进区间的端点? 3、如何推进区间的端点? 4、何时结束区间的枚举? 尺取法通常适用于选取区间有一定规律,或者说所选取的区间有一定的变化趋势的情况,通俗地说,在对所选取区间进行判断之后,我们可以明确如何进一步有方向地推进区间端点以求解满足条件的区间,如果已经判断了目前所选取的区间,但却无法确定所要求解的区间如何进一步得到根据其端点得到,那么尺取法便是不可行的。首先,明确题目所需要求解的量之后,区间左右端点一般从最整个数组的起点开始,之后判断区间是否符合条件在根据实际情况变化区间的端点求解答案。

例子:题意:给定一个序列,使得其和大于或等于S,求最短的子序列长度。

思路:序列都是正数,如果一个区间和大于等于S,那么就不需要往后再推进,因为其和也比大于S,序列还长了,所以这个时候左端点移动以进一步找到最短的区间,如果右端点到了区间末尾其和还不大于等于S,结束区间的枚举。

#include<bits/stdc++.h>
#define maxn 100005
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;

ll a[maxn];

int main(){
	int t;
	cin>>t;
	while(t--){
		int n,s;
		cin>>n>>s;
		for(int i=0;i<n;i++) scanf("%d",a+i);
		int st = 0;
		int en = 0;
		int ans = inf;
		int  sum = 0;
		while(1){
			while(en<n && sum<s) sum += a[en++];//前缀处理
			if(sum < s) break;
			ans = min(ans,en-st);
			sum -= a[st++]; 
		}
		if(ans = inf)  ans = 0;
		printf("%dn",ans);
	}
	return 0;
} 

例子2 一本书有 P 页,每页都有个知识点a[i],知识点可能重复,求包含所有知识点的最少的页数。

思路:跟上一个题差不多,从头开始如果一个子区间满足条件,那么区间推进到该处时,右端点固定,左端点向右移动以求得到最短的子区间。只是需要存储所在区间知识点的数量,那么使用map进行映射以快速的判断是否所选取的页数是否覆盖了所有的知识点

#include <bits/stdc++.h>

#define INF 0x3f3f3f3f
using namespace std;
const int MAXN = 1000010;
int a[MAXN];
map<int, int> hashTable; //用于保存已出现的知识点以及其出现的次数
set<int> temp;
 
int main()
{
 
    int n;
    scanf("%d",&n);
    for(int i = 1;i <= n;++i){
        scanf("%d",&a[i]);
        temp.insert(a[i]); 
    }
    int len = temp.size();  //set容器自动有序并去重,可以利用这一性质来得到不同的知识点总数量
    temp.clear();
    
    int left = 1, right = 1, num = 0, ans = INF;
    
    while(1){
        while(right <= n && num < len){  //此循环仅执行一次,用于判断子序列右端点的最小位置
            if(hashTable[a[right++]]++ == 0) //有没有出现过,先判断后加! 
                ++num;//数目加一 
        }
        if(num < len)// 此分支有两个用处,一是若上面那个循环完成后知识点总数不足要求,就退出循环(这个题不需要)。二是求最大左端点
            break;
            
        ans = min(ans, right - left); //【left,right) 区间左闭右开
        if(--hashTable[a[left++]] == 0) //如果去掉最右边元素,不同的
            --num;
    }
    printf("%dn",ans);
    return 0;
}

例子3 题意:给定一个数组和一个值t,求一个子区间使得其和的绝对值与t的差值最小,如果存在多个,任意解都可行。

#include <cstdio>
#include <algorithm>
#include <cstring>
#define INF 0x3f3f3f3f
#define LL long long
#define MAX 100010
using namespace std;
 
typedef pair<LL, int> p;
LL a[MAX], t, ans, tmp, b;
int n, k, l, u, st, en;
p sum[MAX];
 
LL myabs(LL x)
{
    return x>=0? x:-x;
}
 
int main()
{
    while (scanf("%d %d", &n, &k), n+k){
        sum[0] = p(0, 0);
        for (int i = 1; i <= n; i++){
            scanf("%I64d", a+i);
            sum[i] = p(sum[i-1].first+a[i], i);
        }
        sort(sum, sum+1+n);
        while (k--){
            scanf("%I64d", &t);
            tmp = INF; st = 0, en = 1;
            while(en <= n){
                b = sum[en].first-sum[st].first;
                if(myabs(t-b) < tmp){
                    tmp = myabs(t-b);
                    ans = b;
                    l = sum[st].second; u = sum[en].second;
                }
                if(b > t) st++;
                else if(b < t) en++;
                else break;
                if(st == en) en++;
            }
            if (u < l) swap(u, l);
            printf("%I64d %d %dn", ans, l+1, u);
        }
    }
    return 0;
}

题意:找到某一个区间使得区间内的数的和/平方和等于某一给定值k。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <utility>
#include <queue>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
 
/*
* 提示:该行代码过长,系统自动注释不进行高亮。一键复制会移除系统注释 
* int prime[] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,1597,1601,1607,1609,1613,1619,1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,1973,1979,1987,1993,1997,1999,2003,2011,2017,2027,2029,2039,2053,2063,2069,2081,2083,2087,2089,2099,2111,2113,2129,2131,2137,2141,2143,2153,2161,2179,2203,2207,2213,2221,2237,2239,2243,2251,2267,2269,2273,2281,2287,2293,2297,2309,2311,2333,2339,2341,2347,2351,2357,2371,2377,2381,2383,2389,2393,2399,2411,2417,2423,2437,2441,2447,2459,2467,2473,2477,2503,2521,2531,2539,2543,2549,2551,2557,2579,2591,2593,2609,2617,2621,2633,2647,2657,2659,2663,2671,2677,2683,2687,2689,2693,2699,2707,2711,2713,2719,2729,2731,2741,2749,2753,2767,2777,2789,2791,2797,2801,2803,2819,2833,2837,2843,2851,2857,2861,2879,2887,2897,2903,2909,2917,2927,2939,2953,2957,2963,2969,2971,2999,3001,3011,3019,3023,3037,3041,3049,3061,3067,3079,3083,3089,3109,3119,3121,3137,3163,3167,3169,3181,3187,3191,3203,3209,3217,3221,3229,3251,3253,3257,3259,3271,3299,3301,3307,3313,3319,3323,3329,3331,3343,3347,3359,3361,3371,3373,3389,3391,3407,3413,3433,3449,3457,3461,3463,3467,3469,3491,3499,3511,3517,3527,3529,3533,3539,3541,3547,3557,3559,3571,3581,3583,3593,3607,3613,3617,3623,3631,3637,3643,3659,3671,3673,3677,3691,3697,3701,3709,3719,3727,3733,3739,3761,3767,3769,3779,3793,3797,3803,3821,3823,3833,3847,3851,3853,3863,3877,3881,3889,3907,3911,3917,3919,3923,3929,3931,3943,3947,3967,3989,4001,4003,4007,4013,4019,4021,4027,4049,4051,4057,4073,4079,4091,4093,4099,4111,4127,4129,4133,4139,4153,4157,4159,4177,4201,4211,4217,4219,4229,4231,4241,4243,4253,4259,4261,4271,4273,4283,4289,4297,4327,4337,4339,4349,4357,4363,4373,4391,4397,4409,4421,4423,4441,4447,4451,4457,4463,4481,4483,4493,4507,4513,4517,4519,4523,4547,4549,4561,4567,4583,4591,4597,4603,4621,4637,4639,4643,4649,4651,4657,4663,4673,4679,4691,4703,4721,4723,4729,4733,4751,4759,4783,4787,4789,4793,4799,4801,4813,4817,4831,4861,4871,4877,4889,4903,4909,4919,4931,4933,4937,4943,4951,4957,4967,4969,4973,4987,4993,4999,5003,5009,5011,5021,5023,5039,5051,5059,5077,5081,5087,5099,5101,5107,5113,5119,5147,5153,5167,5171,5179,5189,5197,5209,5227,5231,5233,5237,5261,5273,5279,5281,5297,5303,5309,5323,5333,5347,5351,5381,5387,5393,5399,5407,5413,5417,5419,5431,5437,5441,5443,5449,5471,5477,5479,5483,5501,5503,5507,5519,5521,5527,5531,5557,5563,5569,5573,5581,5591,5623,5639,5641,5647,5651,5653,5657,5659,5669,5683,5689,5693,5701,5711,5717,5737,5741,5743,5749,5779,5783,5791,5801,5807,5813,5821,5827,5839,5843,5849,5851,5857,5861,5867,5869,5879,5881,5897,5903,5923,5927,5939,5953,5981,5987,6007,6011,6029,6037,6043,6047,6053,6067,6073,6079,6089,6091,6101,6113,6121,6131,6133,6143,6151,6163,6173,6197,6199,6203,6211,6217,6221,6229,6247,6257,6263,6269,6271,6277,6287,6299,6301,6311,6317,6323,6329,6337,6343,6353,6359,6361,6367,6373,6379,6389,6397,6421,6427,6449,6451,6469,6473,6481,6491,6521,6529,6547,6551,6553,6563,6569,6571,6577,6581,6599,6607,6619,6637,6653,6659,6661,6673,6679,6689,6691,6701,6703,6709,6719,6733,6737,6761,6763,6779,6781,6791,6793,6803,6823,6827,6829,6833,6841,6857,6863,6869,6871,6883,6899,6907,6911,6917,6947,6949,6959,6961,6967,6971,6977,6983,6991,6997,7001,7013,7019,7027,7039,7043,7057,7069,7079,7103,7109,7121,7127,7129,7151,7159,7177,7187,7193,7207,7211,7213,7219,7229,7237,7243,7247,7253,7283,7297,7307,7309,7321,7331,7333,7349,7351,7369,7393,7411,7417,7433,7451,7457,7459,7477,7481,7487,7489,7499,7507,7517,7523,7529,7537,7541,7547,7549,7559,7561,7573,7577,7583,7589,7591,7603,7607,7621,7639,7643,7649,7669,7673,7681,7687,7691,7699,7703,7717,7723,7727,7741,7753,7757,7759,7789,7793,7817,7823,7829,7841,7853,7867,7873,7877,7879,7883,7901,7907,7919,7927,7933,7937,7949,7951,7963,7993,8009,8011,8017,8039,8053,8059,8069,8081,8087,8089,8093,8101,8111,8117,8123,8147,8161,8167,8171,8179,8191,8209,8219,8221,8231,8233,8237,8243,8263,8269,8273,8287,8291,8293,8297,8311,8317,8329,8353,8363,8369,8377,8387,8389,8419,8423,8429,8431,8443,8447,8461,8467,8501,8513,8521,8527,8537,8539,8543,8563,8573,8581,8597,8599,8609,8623,8627,8629,8641,8647,8663,8669,8677,8681,8689,8693,8699,8707,8713,8719,8731,8737,8741,8747,8753,8761,8779,8783,8803,8807,8819,8821,8831,8837,8839,8849,8861,8863,8867,8887,8893,8923,8929,8933,8941,8951,8963,8969,8971,8999,9001,9007,9011,9013,9029,9041,9043,9049,9059,9067,9091,9103,9109,9127,9133,9137,9151,9157,9161,9173,9181,9187,9199,9203,9209,9221,9227,9239,9241,9257,9277,9281,9283,9293,9311,9319,9323,9337,9341,9343,9349,9371,9377,9391,9397,9403,9413,9419,9421,9431,9433,9437,9439,9461,9463,9467,9473,9479,9491,9497,9511,9521,9533,9539,9547,9551,9587,9601,9613,9619,9623,9629,9631,9643,9649,9661,9677,9679,9689,9697,9719,9721,9733,9739,9743,9749,9767,9769,9781,9787,9791,9803,9811,9817,9829,9833,9839,9851,9857,9859,9871,9883,9887,9901,9907,9923,9929,9931,9941,9949,9967,9973};
*/
 
int main()
{
    int n;
    while (scanf("%d", &n), n){
        int ans, st, en, sum;
        st = en = ans = sum = 0;
        while (1){
            if (sum == n) ans++;
            if (sum >= n) sum -= prime[st++];
            else{
                if (prime[en] <= n) sum += prime[en++];
                else break;
            }
        }
        printf("%dn", ans);
    }
}

总结:尺取法的模型便是这样:根据区间的特征交替推进左右端点求解问题,其高效的原因在于避免了大量的无效枚举,其区间枚举都是根据区间特征有方向的枚举,如果胡乱使用尺取法的话会使得枚举量减少,因而很大可能会错误,所以关键的一步是进行问题的分析!