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 = 3100;
int head[maxn],uf[maxn],dfn[maxn],low[maxn],clk,tot,ret;
struct arc {
    int to,next;
    arc(int x = 0,int z = -1) {
        to = x;
        next = z;
    }
} e[1000010];
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 = 0; i < maxn; ++i) {
            uf[i] = i;
            g[i].clear();
        }
        for(int i = 1,tmp; i < n; ++i) {
            for(int j = i + 1; j <= n; ++j) {
                scanf("%d",&tmp);
                g[tmp].push_back(PII(i,j));
            }
        }
        for(int i = 1; i < maxn; ++i) {
            memset(dfn,0,sizeof dfn);
            memset(head,-1,sizeof head);
            tot = 0;
            for(int j = g[i].size()-1; j >= 0; --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 = 1; j <= n; ++j)
                if(!dfn[j]) tarjan(j,-1);
            for(int j = g[i].size()-1; j >= 0; --j) {
                int u = Find(g[i][j].first);
                int v = Find(g[i][j].second);
                if(u != v) uf[u] = v;
            }
        }
        printf("%d
",ret);
    }
    return 0;
}