【BZOJ1016】[JSOI2008]最小生成树计数【最小生成树】【搜索】

【题目链接】

参考了【hzwer的题解】orz

要利用最小生成树的性质：对于所有的最小生成树，边权相等的边出现次数都相同。

/* Footprints In The Blood Soaked Snow */
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 105, maxm = 1005, p = 31011;

int n, m, fa[maxn], tot, sum, ans;

struct _edge {
int u, v, w;
} g[maxm];

struct _data {
int cnt, l, r;
} e[maxm];

inline int iread() {
int f = 1, x = 0; char ch = getchar();
for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
return f * x;
}

inline int find(int x) {
return fa[x] == x ? x : find(fa[x]);
}

inline bool cmp(_edge a, _edge b) {
return a.w < b.w;
}

inline void dfs(int now, int id, int used) {
if(now == e[id].r + 1) {
if(used == e[id].cnt) sum++;
return;
}
int u = find(g[now].u), v = find(g[now].v);
if(u != v) {
fa[u] = v;
dfs(now + 1, id, used + 1);
fa[u] = u; fa[v] = v;
}
dfs(now + 1, id, used);
}

int main() {
n = iread(); m = iread();
for(int i = 1; i <= m; i++) {
int u = iread(), v = iread(), w = iread();
g[i] = (_edge){u, v, w};
}

for(int i = 1; i <= n; i++) fa[i] = i;
sort(g + 1, g + 1 + m, cmp);

int flag = 0;
for(int i = 1; i <= m; i++) {
if(g[i - 1].w != g[i].w) {
e[tot].r = i - 1; ++tot;
e[tot].l = i;
}
int u = find(g[i].u), v = find(g[i].v);
if(u != v) {
e[tot].cnt++;
fa[u] = v;
flag++;
}
}
e[tot].r = m;

if(flag != n - 1) {
printf("0\n");
return 0;
}

for(int i = 1; i <= n; i++) fa[i] = i;
ans = 1;
for(int i = 1; i <= tot; i++) {
if(e[i].cnt) {
sum = 0;
dfs(e[i].l, i, 0);
} else sum = 1;
ans = ans * sum % p;
for(int j = e[i].l; j <= e[i].r; j++) {
int u = find(g[j].u), v = find(g[j].v);
fa[u] = v;
}
}

printf("%d\n", ans);
return 0;
}