【LOJ】#2069. 「SDOI2016」齿轮

题解

我一开始还努力想这道题是不是有坑，被SDOI折磨到我觉得不能有那么水的题在……

就是带权并查集维护一下两点间距离，如果新加一条边两个点在同一集合，看看已有的路径和新加的路径是否相等

乘积可以在模意义下维护，多随机几个模数就行

代码

#include <bits/stdc++.h>
#define enter putchar('
')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define MAXN 1000005
#define mo 999999137
#define pb push_back
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 + c - '0';
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
int T;
int Mod[5] = {1926365417,1514229737,1066816031,1224683249,1091059271};
int N,M;
int u[10005],v[10005],x[10005],y[10005],inv[105];
int fa[1005],dis[1005];
int MOD;
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
        if(c & 1) res = mul(res,t);
        t = mul(t,t);
        c >>= 1;
    }
    return res;
}
int getfa(int u) {
    if(fa[u] == u) return u;
    else {
        int res = getfa(fa[u]);
        dis[u] = mul(dis[fa[u]],dis[u]);
        fa[u] = res;
        return res;
    }
}
bool check(int c) {
    MOD = c;
    inv[1] = 1;
    for(int i = 2 ; i <= 100 ; ++i) {
        inv[i] = mul(inv[MOD % i],MOD - MOD / i);
    }
    for(int i = 1 ; i <= N ; ++i) {
        fa[i] = i;dis[i] = 1;
    }
    for(int i = 1 ; i <= M ; ++i) {
        int up = x[i],down = inv[abs(y[i])];
        if(x[i] < 0) up = MOD - up;
        if(y[i] < 0) down = MOD - down;
        if(getfa(u[i]) == getfa(v[i])) {
            int r = mul(fpow(dis[v[i]],MOD - 2),dis[u[i]]);
            if(r != mul(up,down)) return false;
        }
        else {
            int r = mul(mul(up,down),fpow(dis[u[i]],MOD - 2));
            int t = getfa(u[i]);
            fa[t] = v[i];dis[t] = r;
        }
    }
    return true;
}
bool Solve() {
    read(N);read(M);
    for(int i = 1 ; i <= M ; ++i) {
        read(u[i]);read(v[i]);read(x[i]);read(y[i]);
    }
    for(int i = 0 ; i <= 4 ; ++i) {
        if(!check(Mod[i])) return false;
    }
    return true;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(T);
    for(int i = 1 ; i <= T ; ++i) {
        printf("Case #%d: ",i);
        if(Solve()) puts("Yes");
        else puts("No");
    }
    return 0;
}