图论BFS

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。

相关推荐
拾光拾趣录5 分钟前
虚拟DOM
前端·vue.js·dom
爱学习的茄子5 分钟前
JavaScript事件循环深度解析:理解异步执行的本质
前端·javascript·面试
1024小神6 分钟前
cocos游戏开发中多角色碰撞,物理反弹后改变方向的实现逻辑
前端·javascript
摆烂为不摆烂10 分钟前
😁深入JS(五): 一文让你完全理解 hash 与 history 路由,手写前端路由
前端
1024小神11 分钟前
cocos游戏开发中,如何制作一个空气墙
前端·javascript
爱编程的喵11 分钟前
深入理解JavaScript事件循环机制:从同步到异步的完整解析
前端·javascript
202612 分钟前
11. vite打包优化
前端·javascript·vite
算家计算14 分钟前
“28项评测23项SOTA——GLM-4.1V-9B-Thinking本地部署教程:10B级视觉语言模型的性能天花板!
人工智能·开源
Codebee14 分钟前
OneCode注解驱动:智能送货单系统的AI原生实现
人工智能·低代码