POJ 1679 The Unique MST

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这个题除了求最小生成树之外，而且要求树的形状唯一。

我在求唯一性的时候是这样做的：如果一个顶点u,他的邻接点有两个分别是v1和v2 ，如果find(v1) == find(v2) && cost(u->v1) == cost(u->v2).那个这个生成树就不唯一。当然需要枚举所有顶点。复杂度高了一些。。

/*********************************************** * Author: fisty * Created Time: 2015/2/28 14:59:19 * File Name   : K.cpp *********************************************** */#include <iostream>#include <cstring>#include <deque>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <vector>#include <cstdio>#include <bitset>#include <algorithm>using namespace std;#define Debug(x) cout << #x << " " << x <<endl#define Memset(x, a) memset(x, a, sizeof(x))const int INF = 0x3f3f3f3f;typedef long long LL;typedef pair<int, int> P;#define FOR(i, a, b) for(int i = a;i < b; i++)#define MAX_N 10010int n, m;int cnt;bool ok;struct edge{    int u;    int v;    int cost;}G[MAX_N];struct node{    int v;    int cost;    node(int _v, int _cost) : v(_v), cost(_cost){}};vector<node> _G[MAX_N];        //判断是否唯一void addedge(int u, int v, int cost){    G[cnt].u = u;    G[cnt].v = v;    G[cnt].cost = cost;    cnt++;}bool cmp(edge e1, edge e2){    return e1.cost < e2.cost;}int par[MAX_N];void init(){    for(int i = 0;i <= n; i++){        par[i] = i;    }}int find(int x){    if(x == par[x]) return x;    return par[x] = find(par[x]);}void unio(int x, int y){    x = find(x);    y = find(y);    if(x != y){        par[x] = y;    }}int kurskal(){    sort(G, G + cnt, cmp);    int ans = 0;    init();    for(int i = 0;i < cnt; i++){        edge e = G[i];        if(find(e.u) != find(e.v)){            unio(e.u, e.v);            ans += e.cost;        }    }    return ans;}bool unique(){        for(int i = 1;i <= n; i++){        for(int j = 0;j < _G[i].size(); j++){            for(int k = 0;k < _G[i].size(); k++){                if(j != k){                    node e1 = _G[i][j], e2 = _G[i][k];                    if(find(e1.v) == find(e2.v) && e1.cost == e2.cost)                        return false;                }            }        }    }    return true;}int main() {    //freopen("in.cpp", "r", stdin);    cin.tie(0);    ios::sync_with_stdio(false);    int t;    cin >> t;    while(t--){        cnt = 0;        Memset(G, 0);        cin >> n >> m;        FOR(i, 1, n+1){            _G[i].clear();        }        FOR(i, 0, m){            int u, v, cost;            cin >> u >> v >> cost;            addedge(u, v, cost);            _G[u].push_back(node(v, cost));            _G[v].push_back(node(u, cost));        }        int ans = kurskal();        ok = unique();        if(!ok){            cout << "Not Unique!" << endl;        }else{            cout << ans << endl;        }    }    return 0;}

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