Be sjy with brush question # #

Notes
"Problem description"
Given a sequence of length m, the subscript number is 1~m. Each element of the sequence is 1~n.
Integer. The cost of defining a sequence is

You can now select two numbers x and Y, and change all x in sequence a to Y. X can be equal to Y.
The minimum possible cost of requesting a sequence.
"Input Format"
Enter the first line containing two integers n and M. The second line contains a m-space-delimited integer, which represents the ordinal
Column A.
"Output Format"
The output line contains an integer that represents the minimum cost of the sequence.
"Sample Input 1"
4 6
1 2 3 4 3 2
"Sample Output 1"
3
"Sample Input 2"
10 5
9 4 3) 8 8
"Sample Output 1"
6
"Sample Interpretation"
In Example 1, the optimal strategy is to change 4 to 3. In Example 2, the optimal strategy is to change 9 to 4.
"Data size and conventions"

For 30% of data, n,m<=100.
For 60% of data, n,m≤2000.
For 100% of data, 1≤n,m≤100,000.

Castle
"Problem description"
Given an undirected connected graph of N-point m-bars, each edge has edge rights. We need to get from the M-bar.
Elect n−1 to form a tree. The shortest path length from point 1th to each node in the original image is Di,
The shortest path length from point 1th to each node in the tree is?? , the tree should be satisfied for any node
I, all have di = Si.
Please find out the number of options to choose n−1.
"Input Format"
The first line of input contains two integers n and M.
Next m line, each line contains three integers u, V and W, describes a connection node U and V and the edge right is
The side of W.
"Output Format"
The output line contains an integer that represents the result of the schema number for the 2^31−1 modulo.
"Sample Input"
3 3
1 2 2
1 3 1
2 3 1
"Sample Output"
2
"Data size and conventions"
For 20% of data, 2<=n<=5,m<=10.
For 50% of data, the number of scenarios that meet the criteria does not exceed 10000.
For 100% of data, 2≤n≤1000,n-1≤m≤n (N-1)/2,1<=w<=100.

#include <iostream>#include<cstdio>#include<cstring>#include<queue>#defineMAXN 1000#defineMAXM 500000#defineMoD 1<<31-1using namespaceStd;typedefLong LongLL;intM,n;intW[MAXN][MAXN], S[MAXN][MAXN], NUM[MAXN], D[MAXN]; LL DIS[MAXN], Sum[maxn], ans (1); Queue<int>que;intMain () {Ios::sync_with_stdio (false); CIN>> N >>m;  for(intI=1; i<=m;i++) {        intU,v,value;cin>>u>>v>>value; W[U][V]= W[v][u] =value; s[u][++num[u]] =v; s[v][++NUM[V]] =u; }     for(intI=1; i<=n;i++) Dis[i] =0x7fffff; Que.push (1);d [1] =1;d is[1] =0;  Do{        inth = Que.front ();d [h] =0;        Que.pop ();  for(intI=1; i<=num[h];i++){            if(Dis[s[h][i] > Dis[h] +W[h][s[h][i]]) {Dis[s[h][i]]= Dis[h] +W[h][s[h][i]]; if(!D[s[h][i]]) {Que.push (S[h][i]), D[s[h][i]=1; }            }        }    } while(!que.empty ()); Que= queue<int>(); memset (d,0,sizeof(d)); Que.push (1), d[1] =1, sum[1] =1;  Do{        inth = Que.front ();d [h] =0;        Que.pop ();  for(intI=1; i<=num[h];i++){            if(Dis[s[h][i]] = = dis[h]+W[h][s[h][i]]) {Sum[s[h][i]]= = mod? Sum[s[h][i]] =1: sum[s[h][i]]++; if(!D[s[h][i]]) {Que.push (S[h][i]), D[s[h][i]=1; }            }        }    } while(!que.empty ());  for(intI=1; i<=n;i++) ans= (Ans*sum[i])%m; cout<< ans <<Endl; return 0;} 

Castle.cpp

Be sjy with brush question # #