Travel Expense(最短路 + 二分）

传送门

写wa了，没开longlong，丢人

n的数据范围很小，那么直接Floyd搞起来，跑完最短路预处理结束之后开始进行判断。
由于给的budget最大为$1e8$，那么如果硬搞是过不去的，直接二分，AC.

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

typedef long long ll;
const int N = 105;
ll n, m, q;
ll s, t, b;
ll maze[N][N], len;
bool check(int x){
	ll r = 1, sum = 0;
	for(int i = 1; i <= len; i ++){
		r *= x;
		sum += r;
		if(sum > b)
			return 0;
	}
    if(sum <= b)
	    return 1;
    else
        return 0;
}
int main(){
    
    cin >> n >> m;
    memset(maze, 0x3f3f3f3f, sizeof maze);
    for(int i = 1; i <= m; i ++){
        int a, b;
        cin >> a >> b;
        maze[a][b] = maze[b][a] = 1;
    }
    for(int k = 1; k <= n; k ++){
        for(int i = 1; i <= n; i ++){
            for(int j = 1; j <= n; j ++){
                if(maze[i][j] > maze[i][k] + maze[k][j]){
                    maze[i][j] = maze[i][k] + maze[k][j];
                    maze[j][i] = maze[i][j];
                }
            }
        }
    }
    cin >> q;
    for(int i = 1; i <= q; i ++){
        
        cin >> s >> t >> b;
        ll l = 0, r = b;
        len = maze[s][t];
        //cout << days << endl;
        while(l < r){
            ll mid = (l + r + 1) >> 1;
            if(check(mid)){
            	l = mid;
			}else
				r = mid - 1;
        }
        cout << l << endl;
    }
    return 0;
}