Moving On（floyd）

题目描述
Firdaws and Fatinah are living in a country with n cities, numbered from 1 to n. Each city has a risk of kidnapping or robbery.
Firdaws’s home locates in the city u, and Fatinah’s home locates in the city v. Now you are asked to find the shortest path from the city u to the city v that does not pass through any other city with the risk of kidnapping or robbery higher than w, a threshold given by Firdaws.

输入
The input contains several test cases, and the first line is a positive integer T indicating the number of test cases which is up to 50.
For each test case, the first line contains two integers n (1≤n≤200) which is the number of cities, and q (1≤q≤2×10⁴) which is the number of queries that will be given.The second line contains nn integers r1, r2, ⋯,rn indicating the risk of kidnapping or robbery in the city 1 to n respectively.Each of the following n lines contains n integers, the j-th one in the i-th line of which, denoted by di,j, is the distance from the city i to the city j.
Each of the following q lines gives an independent query with three integers u, v and w, which are described as above.
We guarantee that 1≤ri≤10⁵, 1≤di,j≤10⁵(i≠j), di,i=0 and di,j=dj,i.Besides, each query satisfies 1≤u,v≤n and 1≤w≤10⁵.

输出
For each test case, output a line containing Case #x: at first, where x is the test case number starting from 1.Each of the following q lines contains an integer indicating the length of the shortest path of the corresponding query.

样例输入
1
3 6
1 2 3
0 1 3
1 0 1
3 1 0
1 1 1
1 2 1
1 3 1
1 1 2
1 2 2
1 3 2

样例输出
Case #1:
0
1
3
0
1
2

思路
对危险程度从小到大进行排序，借助floyd算法递推计算即可

代码实现

  #pragma GCC optimize(3,"Ofast","inline")
#include 

using namespace std;
const int N=305;
typedef long long ll;
const double PI=acos(-1.0);

struct node
{
    int d,no;
}city[N];
int dp[N][N][N];

bool cmp(node a,node b)
{
    return a.dw) break;
            no--;
            printf("%d\n",dp[u][v][no]);
        }
    }
    return 0;
}