POJ 1679 The Unique MST 次小生成树

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题目来源：POJ 1679 The Unique MST

题意：判断最小生成树是否唯一求出次小生成树比较

思路：慢一点的方法就是求出最小生成树每次去掉最小生成树的一条边再求最小生成树比较慢

更好的方法是求出最小生成树后加上一条没有用到的边然后必定出现一条回路去掉回路上权值最大的边做m-(n-1)次

求一次最小生成树然后n^2的时间预处理最小生成树上两点之间最大的边权最后换边 MST-加进去的边的权值-回路上的最大边权值取小

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>using namespace std;const int maxn = 110;struct edge{int u, v, w;edge(){}edge(int u, int v, int w): u(u), v(v), w(w){}}a[100000];int f[maxn];int dp[maxn][maxn];int n, m;bool ok[100000];vector <edge> G[maxn];bool cmp(edge a, edge b){return a.w < b.w;}int find(int x){if(x != f[x])return f[x] = find(f[x]);return f[x];}int MST(){for(int i = 1; i <= n; i++){f[i] = i;G[i].clear();}memset(ok, false, sizeof(ok));sort(a, a+m, cmp);memset(dp, 0, sizeof(dp));int sum = 0;for(int i = 0; i < m; i++){int u = a[i].u;int v = a[i].v;int w = a[i].w;int x = find(u);int y = find(v);if(x != y){sum += w;f[x] = y;G[u].push_back(edge(u, v, w));G[v].push_back(edge(v, u, w));ok[i] = true;}}return sum;}void BFS(int x){queue <edge> Q;Q.push(edge(-1, x, 0));while(!Q.empty()){edge u = Q.front(); Q.pop();for(int i = 0; i < G[u.v].size(); i++){edge v = G[u.v][i];if(v.v == u.u)continue;int temp = max(u.w, v.w);dp[x][v.v] = dp[v.v][x] = max(dp[x][v.v], temp);Q.push(edge(v.u, v.v, temp));}}}int main(){int T;scanf("%d", &T);while(T--){scanf("%d %d", &n, &m);for(int i = 0; i < m; i++)scanf("%d %d %d", &a[i].u, &a[i].v, &a[i].w);int ans = MST();int sum = 999999999;for(int i = 1; i <= n; i++)BFS(i);for(int i = 0; i < m; i++){if(ok[i])continue;int u = a[i].u;int v = a[i].v;int w = a[i].w;sum = min(sum, ans+w-dp[u][v]);}if(ans == sum)puts("Not Unique!");elseprintf("%d\n", ans);}return 0;}

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