题意：给定两个数，表示一个图的点数和边数，让你构造出一个图满足 1- n 的最短路是素数，并且最小生成树也是素数。
析：首先 1 - n 的最短路，非常好解决，直接 1 连 n 就好了，但是素数尽量选小的，选2，3，5，这样比较小的，然后再构造MST，可以给每个边都是 1，然后最后 n-2 连 n-1的时候，保证加起来是素数就好，然后剩下的边随便连，凑够边数就好，但是权值要尽量的大，但不要超过1e9，一开始没看到 1e9，写大了，1e8就够用了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
//#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 1e8;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 10;
const int maxm = 3e5 + 10;
const LL mod = 1e9 + 7LL;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
bool is_prime(int x){
int t = sqrt(x + 0.5);
for(int i = 2; i <= t; ++i)
if(x % i == 0) return false;
return true;
}
int main(){
scanf("%d %d", &n, &m);
if(n == 2){ printf("3 3
1 2 3
"); return 0; }
int sp = 3, mstp = m + 2;
for(int i = m + 2; !is_prime(i); mstp = ++i);
printf("%d %d
", sp, mstp);
printf("%d %d %d
", 1, n, 3);
int det = mstp - n;
m -= 2;
for(int i = 1; i <= n-3; ++i, --m) printf("%d %d 1
", i, i + 1);
printf("%d %d %d
", n-2, n-1, det);
if(m) printf("%d %d %d
", n-1, n, INF);
--m;
for(int i = 1; i <= n && m > 0; ++i)
for(int j = i+2; j <= n && m > 0; ++j)
if(i != 1 || j != n) printf("%d %d %d
", i, j, INF), --m;
return 0;
}