【POJ 1679】 The Unique MST

【题目链接】

点击打开链接

【算法】

先求出图的最小生成树

枚举不在最小生成树上的边，若加入这条边，则形成了一个环，如果在环上且在最小生成树上的权值最大的边等于

这条边的权值，那么，显然最小生成树不唯一

树上倍增可以解决这个问题

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 110
#define MAXM 1000010
#define MAXLOG 20
 
struct Edge
{
        int x,y;
        long long w;
} edge[MAXM];
 
int T,n,m,i;
long long val;
vector< pair<int,long long> > e[MAXN];
bool on_mst[MAXM];
int fa[MAXN],anc[MAXN][MAXLOG],dep[MAXN];
long long mx[MAXN][MAXLOG];
bool not_unique;
 
inline bool cmp(Edge a,Edge b) { return a.w < b.w; }
inline int get_root(int x)
{
        if (fa[x] == x) return x;
        return fa[x] = get_root(fa[x]);
}
inline void kruskal()
{
        int i,x,y,sx,sy;
        long long w;
        for (i = ; i <= n; i++) fa[i] = i;
        for (i = ; i <= m; i++) on_mst[i] = false;
        sort(edge+,edge+m+,cmp);
        for (i = ; i <= m; i++)
        {
                x = edge[i].x;
                y = edge[i].y;
                w = edge[i].w;
                sx = get_root(x);
                sy = get_root(y);
                if (sx != sy)
                {
                        on_mst[i] = true;
                        val += w;
                        fa[sx] = sy;
                        e[x].push_back(make_pair(y,w));
                        e[y].push_back(make_pair(x,w));
                }
        }
}
inline void build(int u)
{
        int i,v;
        for (i = ; i < MAXLOG; i++)
        {
                anc[u][i] = anc[anc[u][i-]][i-];
                mx[u][i] = max(mx[u][i-],mx[anc[u][i-]][i-]);
        }
        for (i = ; i < e[u].size(); i++)
        {
                v = e[u][i].first;
                if (anc[u][] != v)
                {
                        dep[v] = dep[u] + ;
                        anc[v][] = u;
                        mx[v][] = e[u][i].second;
                        build(v);
                }
        }
}
inline long long get(int x,int y)
{
        int i,t;
        long long ans = ;
        if (dep[x] > dep[y]) swap(x,y);
        t = dep[y] - dep[x];
        for (i = ; i < MAXLOG; i++)
        {
                if (t & ( << i))
                {
                        ans = max(ans,mx[y][i]);
                        y = anc[y][i];
                }
        }
        if (x == y) return ans;
        for (i = MAXLOG - ; i >= ; i--)
        {
                if (anc[x][i] != anc[y][i])
                {
                        ans = max(ans,max(mx[x][i],mx[y][i]));
                        x = anc[x][i];
                        y = anc[y][i];
                }
        }
        return max(ans,max(mx[x][],mx[y][]));
}
int main()
{
 
        scanf("%d",&T);
        while (T--)
        {
                scanf("%d%d",&n,&m);
                val = ;
                not_unique = false;
                for (i = ; i <= n; i++)
                {
                        dep[i] = ;
                        e[i].clear();
                        memset(anc[i],,sizeof(anc[i]));
                        memset(mx[i],,sizeof(mx[i]));
                }
                for (i = ; i <= m; i++) scanf("%d%d%lld",&edge[i].x,&edge[i].y,&edge[i].w);
                kruskal();
                build();
                for (i = ; i <= m; i++)
                {
                        if (!on_mst[i])
                                not_unique |= (get(edge[i].x,edge[i].y) == edge[i].w);
                }
                if (not_unique) printf("Not Unique!\n");
                else printf("%lld\n",val);
        }
 
        return ;
 
}