【BZOJ】1016: [JSOI2008]最小生成树计数

题解

考虑kruskal
我们都是从边权最小的边开始取,然后连在一起

那我们选出边权最小的一堆边,然后这个图就分成了很多联通块,把每个联通块内部用矩阵树定理算一下生成树个数,再把联通块缩成一个大点,重复取下一个边权的边进行操作

好想然而不是很好写= =写起来感觉有点麻烦

模数非质数,用long double水一下过掉了

代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <cmath>
#include <bitset>
#include <queue>
#define enter putchar('\n')
#define space putchar(' ')
//#define ivorysi
#define pb push_back
#define mo 974711
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define MAXN 200005
#define eps 1e-12
using namespace std;
typedef long long int64;
typedef long double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 - '0' + c;
    c = getchar();
    }
    res = 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 N,M;
struct node {
    int u,v,c;
    friend bool operator < (const node &a,const node &b) {
    return a.c < b.c;
    }
}E[1005];
int fa[105],id[105],f[105][105],L[105],tot,D[105];
bool vis[105];
db g[105][105];
int64 ans = 1;
int getfa(int x) {
    return x == fa[x] ? x : fa[x] = getfa(fa[x]);
} 
db Guass(int n) {
    db res = 1;
    for(int i = 2 ; i <= n ; ++i) {
    int l = i;
    for(int j = i + 1 ; j <= n ; ++j) {
        if(fabs(g[j][i]) > fabs(g[l][i])) l = j;
    }
    if(i != l) {
        res = -res;
        for(int j = i ; j <= n ; ++j) {
        swap(g[i][j],g[l][j]);
        }
    }
    if(fabs(g[i][i]) == 0) return 0;
    for(int j = i + 1 ; j <= n ; ++j) {
        db t = g[j][i] / g[i][i];
        for(int k = i ; k <= n ; ++k) {
        g[j][k] -= g[i][k] * t;
        }
    }
    }
    for(int k = 2 ; k <= n ; ++k) {
    res = res * g[k][k];
    }
    return res;
}
void dfs(int u,int n) {
    vis[u] = 1;
    L[++tot] = u;
    D[u] = tot;
    for(int i = 1 ; i <= n ; ++i) {
    if(f[u][i]) {
        if(!vis[i]) dfs(i,n);
    }
    }
}
void Process(int l,int r) {
    memset(id,0,sizeof(id));
    memset(f,0,sizeof(f));
    memset(vis,0,sizeof(vis));
    int cnt = 0;
    for(int i = 1 ; i <= N ; ++i) {
    if(!id[getfa(i)]) {
        id[getfa(i)] = ++cnt;
    }
    }
    for(int i = l ; i <= r ; ++i) {
    int s = getfa(E[i].u),t = getfa(E[i].v);
    if(s == t) continue;
    f[id[s]][id[t]]++; f[id[t]][id[s]]++;
    }
    for(int i = 1 ; i <= cnt ; ++i) {
    if(!vis[i]) {
        tot = 0;
        dfs(i,cnt);
        if(tot == 1) continue;
        memset(g,0,sizeof(g));
        for(int j = 1 ; j <= tot ; ++j) {
        int u = L[j];
        for(int k = 1 ; k <= cnt ; ++k) {
            if(f[u][k]) {
            g[j][j] += f[u][k];
            g[j][D[k]] -= f[u][k];
            }
        }
        }
        ans = ans * ((int64)(Guass(tot) + 0.5) % 31011) % 31011;
    }
    }
    for(int i = l ; i <= r ; ++i) {
    fa[getfa(E[i].v)] = getfa(E[i].u);
    }
}
void Solve() {
    read(N);read(M);
    for(int i = 1 ; i <= N ; ++i) fa[i] = i;
    int u,v,c;
    for(int i = 1 ; i <= M ; ++i) {
    read(E[i].u);read(E[i].v);read(E[i].c);
    fa[getfa(E[i].u)] = getfa(E[i].v);
    }
    for(int i = 2 ; i <= N ; ++i) {
    if(getfa(i) != getfa(i - 1)) {
        puts("0");
        return;
    }
    }
    for(int i = 1 ; i <= N ; ++i) fa[i] = i;
    sort(E + 1,E + M + 1);
    v = 0;int st = 0;
    bool flag = 0;
    for(int i = 1 ; i <= M ; ++i) {
    if(E[i].c != v) {
        if(st != 0) Process(st,i - 1);
        st = i;
        v = E[i].c;
        flag = 1;
        for(int j = 2 ; j <= N ; ++j) {
        if(getfa(j) != getfa(j - 1)) {flag = 0;break;}
        }
        if(flag) break;
    }
    }
    if(!flag) Process(st,M);
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}
posted @ 2018-06-22 08:55  sigongzi  阅读(...)  评论(... 编辑 收藏