Bzoj 1016 Jsoi 2008 minimum Spanning tree count kruskal+ search

Topic: Give some sides, find out how many of the minimum spanning tree can be formed.

Idea: The minimum spanning tree has very many theorems ah, I am not very clear. Here is just a simple way to talk about practice. For the various theorems, see here: http://blog.csdn.net/wyfcyx_forever/article/details/40182739

Let's do the minimum spanning tree first. Then record how many edges of each length are in the smallest spanning tree, and then search from small to large to see how many different ways to put each edge right. And then all of them are counted out, and the result is finally.

CODE:

#include <map> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm > #define MAX 2010#define MO 31011using namespace std;map<int,int> g;struct complex{int x,y,len;bool operator &lt ;(Const Complex &a) Const {return Len < A.len;} void Read () {scanf ("%d%d%d", &x,&y,&len);}} Edge[max];int Points,edges;int ones[max];int father[max];int ans = 1;void pretreatment () {for (int i = 1;i <= 1025; ++i) o Nes[i] = ones[i >> 1] + (i&1);} int Find (int x) {if (!father[x] | | father[x] = = x) return father[x] = X;return Father[x] = find (Father[x]);} BOOL Kruskal () {int cnt = 0;for (int i = 1;i <= edges; ++i) {int fx = find (edge[i].x); int fy = find (EDGE[I].Y); if (FX! = f Y) {father[fy] = FX; g[edge[i].len]++;cnt++;}} if (CNT < points-1) return False;return true;} void DFS (int pos) {if (pos > Edges) return; int st = pos,ed = Pos,re = 0;int cnt = g[edge[st].len];if (!cnt) {DFS (ed + 1); r Eturn;} while (edge[ed + 1].len = = Edge[st].len) ++ed;for (int i = 0;i < (1 << (ed-st + 1)), ++i) if (ones[i] = = cnt) {memset (father,0,sizeof (father)); int temp = I;for ( int j = St; Temp Temp >>= 1,++j) if (temp&1) {int fx = find (edge[j].x); int fy = find (EDGE[J].Y); if (fx = = FY) break;father[fy] = FX;} if (!temp) ++re;} Ans = (ans * Re)% Mo;memset (father,0,sizeof (father)); for (int i = st; I <= ed; ++i) {int fx = Find (edge[i].x); int fy = Find (EDGE[I].Y); if (fx = = FY) continue;father[fy] = FX;} for (int i = ed + 1;i <= edges; ++i) edge[i].x = Find (edge[i].x), edge[i].y = Find (EDGE[I].Y);D FS (ed + 1);} int main () {pretreatment (); Cin >> points >> edges;for (int i = 1;i <= edges; ++i) edge[i]. Read (); Sort (edge + 1,edge + edges + 1); Memset (father,0,sizeof (father)); Kruskal ()) {cout << 0 << endl;return 0;} DFS (1); cout << ans << endl;return 0;}

Bzoj 1016 Jsoi 2008 minimum Spanning tree count kruskal+ search