2018徐州现场赛A

题目链接：http://codeforces.com/gym/102012/problem/A

题目给出的算法是通过异或和位移 , 每次生成不同的32位, 周期为2的32次方减1。题目的总边数太少，跑不出重边，所以有最小生成树的话就只有一个

#include<iostream>
#include<algorithm>
using namespace std;
unsigned long long k1,k2;
#define inf 0x3f3f3f3f
#define mod 1000000007
int tot;
unsigned long long xo()
{
    unsigned long long k3=k1,k4=k2;
    k1=k4;
    k3^=k3<<23;
    k2=k3^k4^(k3>>17)^(k4>>26);
    return k2+k4;
}
struct edge{
    int u,v;
    unsigned long long w;
    bool operator <(const edge &a)const{
        return w<a.w;
    }
}e[100010];
int n,m,fa[100010],cnt;
unsigned long long ans;
void add(int u,int v,unsigned long long w)
{
    e[++cnt].u=u;
    e[cnt].v=v;
    e[cnt].w=w;
}
void gen()
{
    int u,v;
    unsigned long long w;
    scanf("%d%d%llu%llu",&n,&m,&k1,&k2);
    cnt=0;
    for(int i=1;i<=m;i++)
    {
        u=xo()%n+1;
        v=xo()%n+1;
        w=xo();
        add(u,v,w);
    }
}
int _find(int u)
{
    if(fa[u]==u)return u;
    return fa[u]=_find(fa[u]);//路径压缩
}
unsigned long long solve()
{
    tot=1;
    sort(e+1,e+1+cnt);
    int u,v,x,y;
    unsigned long long w;
    ans=0;
    for(int i=1;i<=n;i++)
    fa[i]=i;
    for(int i=1;i<=cnt;i++)
    {
        u=e[i].u;v=e[i].v;w=e[i].w;
        x=_find(u);y=_find(v);
        if(x!=y)
        {
            ans=(ans+w)%mod;
            tot++;
            fa[y]=x;
            if(tot==n)return ans;
        }
    }
    return 0;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        gen();
        ans=inf;
        printf("%llu\n",solve());
    }
    return 0;
}

原文地址：https://www.cnblogs.com/chen99/p/11707579.html