cf916CJamie and Interesting Graph

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/*
构造题就要找到一些特殊信息来做
最小生成树总和质数, 1到n最短路路径和质数
这两个信息我们可以整合成一个信息
最小生成树是1到n的一条链

其他的边随便加上极大值就好了


*/

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<set>
#include<ctime>
#include<iostream>
#define M 410
#define ll long long
using namespace std;
int f[M][M], d[M][M];
const int inf = 998244353;
int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == -) f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c -0;
    return nm * f;
}
set<pair<int,int> >st;

int main() {
    srand(time(0));
    int n = read(), m = read();    
    cout << inf << " " << inf << "
";
    int zz = inf;
    for(int i = 1; i < n - 1; i ++) {
        cout << i << " " << i + 1 << " " << "1" << "
";
        st.insert(make_pair(i, i + 1));
        zz--;
    }
    cout << n - 1 << " " << n << " " << zz << "
";
    st.insert(make_pair(n - 1, n));
    m -= n - 1;
    for(int i = 1; i <= m; i++) {
        while(1) {
            int vi = rand() % n + 1, vj = rand() % n + 1;
            if(vi == vj) continue;
            if(vi > vj)swap(vi, vj);
            if(st.find(make_pair(vi, vj)) != st.end()) continue;
            else {
                st.insert(make_pair(vi, vj));
                cout << vi << " " << vj  << " " << inf + 1 << "
";
                break;
            }
        }
    }
    return 0;
}

 

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