HDU 5483 Nux Walpurgis

Nux Walpurgis

Time Limit: 8000ms

Memory Limit: 131072KB

This problem will be judged on HDU. Original ID: 5483
64-bit integer IO format: %I64d      Java class name: Main

Given a weighted undirected graph, how many edges must be on the minimum spanning tree of this graph?

 

Input

The first line of the input is a integer T, meaning that there are T test cases.

Every test cases begin with a integer n ,which is the number of vertexes of this graph.

Then n−1 lines follow, the ith line contain n−i integers, the jth number w in this line represents the weight between vertex i and vertex i+j.

1≤T≤20.

1≤n,w≤3,000.

Output

For every test case output the number of edges must be on the minimum spanning tree of this graph.

 

Sample Input

2
3
1 1
1
4
2 2 3
2 2
3

Sample Output

0
1

Source

BestCoder Round #57 (div.1) ($)

 

解题：题目还是挺难想的，求图的最小生成树的必要边的数量，学习的是某位大神的解法

 #include <bits/stdc++.h>
 using namespace std;
 using PII = pair<int,int>;
 const int maxn = ;
 int head[maxn],uf[maxn],dfn[maxn],low[maxn],clk,tot,ret;
 struct arc {
     int to,next;
     arc(int x = ,int z = -) {
         to = x;
         next = z;
     }
 } e[];
 vector<PII>g[maxn];
 void add(int u,int v) {
     e[tot] = arc(v,head[u]);
     head[u] = tot++;
 }
 int Find(int x) {
     if(x != uf[x]) uf[x] = Find(uf[x]);
     return uf[x];
 }
 void tarjan(int u,int fa) {
     dfn[u] = low[u] = ++clk;
     bool flag = true;
     for(int i = head[u]; ~i; i = e[i].next) {
         if(e[i].to == fa && flag) {
             flag = false;
             continue;
         }
         if(!dfn[e[i].to]) {
             tarjan(e[i].to,u);
             low[u] = min(low[u],low[e[i].to]);
             if(low[e[i].to] > dfn[u]) ++ret;
         } else low[u] = min(low[u],dfn[e[i].to]);
     }
 }
 int main() {
     int kase,n;
     scanf("%d",&kase);
     while(kase--) {
         scanf("%d",&n);
         for(int i = ret = ; i < maxn; ++i) {
             uf[i] = i;
             g[i].clear();
         }
         for(int i = ,tmp; i < n; ++i) {
             for(int j = i + ; j <= n; ++j) {
                 scanf("%d",&tmp);
                 g[tmp].push_back(PII(i,j));
             }
         }
         for(int i = ; i < maxn; ++i) {
             memset(dfn,,sizeof dfn);
             memset(head,-,sizeof head);
             tot = ;
             for(int j = g[i].size()-; j >= ; --j) {
                 int u = Find(g[i][j].first);
                 int v = Find(g[i][j].second);
                 if(u != v) {
                     add(u,v);
                     add(v,u);
                 }
             }
             for(int j = ; j <= n; ++j)
                 if(!dfn[j]) tarjan(j,-);
             for(int j = g[i].size()-; j >= ; --j) {
                 int u = Find(g[i][j].first);
                 int v = Find(g[i][j].second);
                 if(u != v) uf[u] = v;
             }
         }
         printf("%d\n",ret);
     }
     return ;
 }