PAT--1044. Shopping in Mars

最新推荐文章于 2020-07-04 17:57:38 发布

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本文介绍了一种用于解决火星购物支付问题的独特算法。该算法通过双指针技术在钻石链上寻找能够恰好匹配支付金额的钻石组合，如果无法精确匹配，则提供损失最小化的支付方案。

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Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken off the chain one by one. Once a diamond is off the chain, it cannot be taken back. For example, if we have a chain of 8 diamonds with values M$3, 2, 1, 5, 4, 6, 8, 7, and we must pay M$15. We may have 3 options:

Cut the chain between 4 and 6, and take off the diamonds from the position 1 to 5 (with values 3+2+1+5+4=15).
Cut before 5 or after 6, and take off the diamonds from the position 4 to 6 (with values 5+4+6=15).
Cut before 8, and take off the diamonds from the position 7 to 8 (with values 8+7=15).
Now given the chain of diamond values and the amount that a customer has to pay, you are supposed to list all the paying options for the customer.

If it is impossible to pay the exact amount, you must suggest solutions with minimum lost.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 numbers: N (<=105), the total number of diamonds on the chain, and M (<=108), the amount that the customer has to pay. Then the next line contains N positive numbers D1 … DN (Di<=103 for all i=1, …, N) which are the values of the diamonds. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print “i-j” in a line for each pair of i <= j such that Di + … + Dj = M. Note that if there are more than one solution, all the solutions must be printed in increasing order of i.

If there is no solution, output “i-j” for pairs of i <= j such that Di + … + Dj > M with (Di + … + Dj - M) minimized. Again all the solutions must be printed in increasing order of i.

It is guaranteed that the total value of diamonds is sufficient to pay the given amount.

Sample Input 1:

16 15
3 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13

Sample Output 1:

1-5
4-6
7-8
11-11

Sample Input 2:

5 13
2 4 5 7 9

Sample Output 2:

2-4
4-5

题解

用两个指针i, j维护一段区间[i, j]，区间元素之和为sum:
sum < m，则 j++；否则 sum - a[i++]。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 100010;
const int inf  = 0x3f3f3f3f;
int a[maxn];
int n, m;
vector<pair<int, int>> p;

int main(){
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
#endif // ONLINE_JUDGE

    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; ++i) scanf("%d", a + i);

    int i = 1, j = 1;
    int sum = 0;
    int curval = inf;
    while(i <= j){
        while(j <= n && sum < m){
            sum += a[j++];
        }

        int val = sum - m;
        if(val >= 0 && val < curval){
            curval = val;
            p.clear();
            p.push_back(make_pair(i, j - 1));
        }else if(val == curval){
            p.push_back(make_pair(i, j - 1));
        }

        sum -= a[i++];
    }

    for(i = 0; i < p.size(); ++i){
        printf("%d-%d\n", p[i].first, p[i].second);
    }

    return 0;
}