图论BFS - 技术栈

  

D1. The Endspeaker (Easy Version)

time limit per test

2 seconds

memory limit per test

256 megabytes

  

This is the easy version of this problem. The only difference is that you only need to output the minimum total cost of operations in this version. You must solve both versions to be able to hack.

You're given an array aa of length nn, and an array bb of length mm (bi>bi+1bi>bi+1 for all 1≤i<m1≤i<m). Initially, the value of kk is 11. Your aim is to make the array aa empty by performing one of these two operations repeatedly:

Type 11 --- If the value of kk is less than mm and the array aa is not empty, you can increase the value of kk by 11. This does not incur any cost.
Type 22 --- You remove a non-empty prefix of array aa, such that its sum does not exceed bkbk. This incurs a cost of m−km−k.

You need to minimize the total cost of the operations to make array aa empty. If it's impossible to do this through any sequence of operations, output −1−1. Otherwise, output the minimum total cost of the operations.

  

Input

Each test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤10001≤t≤1000). The description of the test cases follows.

The first line of each test case contains two integers nn and mm (1≤n,m≤3⋅1051≤n,m≤3⋅105, 1≤n⋅m≤3⋅1051≤n⋅m≤3⋅105).

The second line of each test case contains nn integers a1,a2,...,ana1,a2,...,an (1≤ai≤1091≤ai≤109).

The third line of each test case contains mm integers b1,b2,...,bmb1,b2,...,bm (1≤bi≤1091≤bi≤109).

It is also guaranteed that bi>bi+1bi>bi+1 for all 1≤i<m1≤i<m.

It is guaranteed that the sum of n⋅mn⋅m over all test cases does not exceed 3⋅1053⋅105.

  

Output

For each test case, if it's possible to make aa empty, then output the minimum total cost of the operations.

If there is no possible sequence of operations which makes aa empty, then output a single integer −1−1.

cpp 复制代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define debug(x) cout << "[debug]" << " = " << x << '\n'
typedef std::pair<int,int> pii;
const int N = 15 + 5, MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const ll INFF = 0x3f3f3f3f3f3f3f3f;
int dx[] = {0, -1, 0, 1}, dy[] = {-1, 0, 1, 0};

void solve() {
    ll n;
    cin >> n;
    vector<ll> a(n+1);
    map<ll, int> visit;
    map <ll, vector<ll>> mp;
    for(ll i = 1; i <= n; i ++){
        cin >> a[i];
        ll u = a[i] + i - 1;
        ll v = u + i - 1;
        mp[u].push_back (v);
    }
    ll max = n;
    queue<ll> q;
    for(auto i: mp[n]) {
        q.push(i);
        if(i > max) max = i;
        visit[i] =1;
    }
    while (!q.empty()) {
        ll j = q.front();
        q.pop();
        for(auto i: mp[j]) {
            if(visit[i] == 0) {
                q.push(i);
                if(i > max) max = i;
                visit[i] =1;
            }
        }
    }
    cout << max << '\n';
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0); std::cout.tie(0);
    //std::cout << std::fixed << std::setprecision(2);
    int T = 1;
    std::cin >> T;
    while(T --) solve();
    return 0;
}

div2 的c题使用数组转化成图论问题，找出节点最大值，因为其中的数值较大，使用map存储路径和访问节点，遍历时间复杂度为NlogN。