_bzoj1016 [JSOI2008]最小生成树计数【生成树】

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=1016

其实原题不叫这个的,而且原题是有一个背景故事的。。。

首先,容易得知,一个最小生成树不管是什么样的,同一种长度的边出现的次数是一样的。而且由Kruscal的过程,若当前节点对整个图减少一个联通块有贡献,则加这条边,也就是说在比当前边短的边已经全部考虑完后,与当前边一样长的所有边对整个图的连通性的贡献是一样的。则,若最小生成树中与当前边一样长的边出现了x次,共有tot条,那么暴力枚举C(tot, x)次,若它们不构成环,则是合法的。最后对每种长度的边乘法原理一下就ok咯~

注意题目陷阱:原图有可能没有生成树,此时输出0!

#include <cstdio>
#include <algorithm>
#include <cstring>
 
const int maxn = 105, maxm = 1005, mod = 31011;
 
int n, m, fa[maxn], num[maxm], cnt, t_fa[15][maxn], idx, fu, fv, ans, t_ans;
int head[maxm], to[maxm], next[maxm], lb, max_step;
struct Edge {
    int u, v, ww, w;
} a[maxm];
 
bool cmp(const Edge & aa, const Edge & ss) {
    return aa.ww < ss.ww;
}
int getfa(int * ff, int aa) {
    return ff[aa] == aa? aa: ff[aa] = getfa(ff, ff[aa]);
}
inline void ist(int aa, int ss) {
    to[lb] = ss;
    next[lb] = head[aa];
    head[aa] = lb;
    ++lb;
}
void dfs(int step, int now) {
    if (step == max_step) {
        ++t_ans;
        return;
    }
    for (int j = next[now]; j != -1; j = next[j]) {
        fu = getfa(t_fa[step - 1], a[to[j]].u);
        fv = getfa(t_fa[step - 1], a[to[j]].v);
        if (fu == fv) {
            continue;
        }
        memcpy(t_fa[step], t_fa[step - 1], sizeof fa);
        t_fa[step][fu] = fv;
        dfs(step + 1, j);
    }
}
 
int main(void) {
    //freopen("in.txt", "r", stdin);
    memset(head, -1, sizeof head);
    memset(next, -1, sizeof next);
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) {
        fa[i] = i;
    }
    for (int i = 0; i < m; ++i) {
        scanf("%d%d%d", &a[i].u, &a[i].v, &a[i].ww);
    }
    std::sort(a, a + m, cmp);
    a[0].w = 1;
    idx = 1;
    ist(1, 0);
    for (int i = 1; i < m; ++i) {
        if (a[i].ww == a[i - 1].ww) {
            a[i].w = idx;
        }
        else {
            a[i].w = ++idx;
        }
        ist(a[i].w, i);
    }
     
    memcpy(t_fa[0], fa, sizeof fa);
    for (int i = 0; i < m; ++i) {
        fu = getfa(fa, a[i].u);
        fv = getfa(fa, a[i].v);
        if (fu != fv) {
            fa[fu] = fv;
            ++num[a[i].w];
            if (++cnt == n - 1) {
                break;
            }
        }
    }
    if (cnt < n - 1) {
        puts("0");
        return 0;
    }
    memcpy(fa, t_fa[0], sizeof fa);
     
    ans = 1;
    for (int i = 1; i <= n; ++i) {
        fa[i] = i;
    }
    for (int i = 1; i <= idx; ++i) {
        if (!num[i]) {
            continue;
        }
        max_step = num[i];
        t_ans = 0;
        for (int j = head[i]; j != -1; j = next[j]) {
            fu = getfa(fa, a[to[j]].u);
            fv = getfa(fa, a[to[j]].v);
            if (fu == fv) {
                continue;
            }
            memcpy(t_fa[0], fa, sizeof fa);
            t_fa[0][fu] = fv;
            dfs(1, j);
        }
        for (int j = head[i]; j != -1; j = next[j]) {
            fu = getfa(fa, a[to[j]].u);
            fv = getfa(fa, a[to[j]].v);
            if (fu != fv) {
                fa[fu] = fv;
            }
        }
        ans = ans * t_ans % mod;
    }
    printf("%d\n", ans);
    return 0;
}

  

posted @ 2016-12-15 20:49  ciao_sora  阅读(...)  评论(... 编辑 收藏