AtCoder-abc257_d Jumping Takahashi 2

Jumping Takahashi 2

dijkstra 或 Floyd 或 二分

dijkstra 枚举从每个点出发到所有点的最大代价的最小值

同理，Floyd 跑完之后再找这个最小值

二分答案，通过 bfs 判断是否有个点能跑遍历所有的点

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;
typedef long long ll;
#define pii pair<double, int>
const int maxn = 210;
const double eps = 1e-8;
ll gra[maxn][maxn];
int vis[maxn];

struct node
{
    ll x, y, val;
    node(){}
    node(ll _x, ll _y, ll _v){x = _x; y = _y; val = _v;}
    bool operator < (const node& a) const
    {
        return a.val < val;
    }
}num[maxn];

ll dis(node& a, node& b)
{
    return abs(a.x - b.x) + abs(a.y - b.y);
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    for(int i=1; i<=n; i++)
        cin >> num[i].x >> num[i].y >> num[i].val;

    for(int i=1; i<=n; i++)
    {
        for(int j=i+1; j<=n; j++)
        {
            ll s = dis(num[i], num[j]);
            gra[i][j] = (s + num[i].val - 1) / num[i].val;
            gra[j][i] = (s + num[j].val - 1) / num[j].val;
        }
    }
    ll ans = 1e17 + 10;
    for(int i=1; i<=n; i++)
    {
        for(int j=j=1; j<=n; j++) vis[j] = 0;
        priority_queue<node>q;
        q.push(node(i, i, 0));
        ll temp = 0;
        int cnt = n;
        while(q.size() && cnt)
        {
            node now = q.top();
            q.pop();
            if(vis[now.y]) continue;
            cnt--;
            vis[now.y] = 1;
            temp = max(temp, now.val);
            for(int j=1; j<=n; j++)
                if(vis[j] == 0)
                    q.push(node(now.y, j, gra[now.y][j]));
        }
        ans = min(ans, temp);
    }
    cout << ans << endl;
    return 0;
}