POJ1679-The Unique MST（次小生成树+最小瓶颈路 ）

题目链接

http://poj.org/problem?id=1679

题意

给定求一张图，判断它的最小生成树是否唯一

思路

只需求出该图的次小生成树，并且判断最小生成树和次小生成树权值和是否相同即可
那么问题就转化成了求次小生成树：先求出最小生成树，然后枚举剩下的m - n +１条边， 将其中一条边(u, v)加入MST中，这样必定u->v之间会形成回路，根据最小生成树的回路性质，回路中权值最大的那条边必定不在最小生成树中，因此我们只需删除MST中u->v权值最大的那条边，即u->v之间的最小瓶颈路
于是可以考虑预处理出任意一对结点之间的最小瓶颈路，可以知道任意一对结点u->v的最小瓶颈路就为MST上u->v的路径。因此，只需求出原图的MST，并且DFS将其转化为有根树并求任意一对结点之间的最小瓶颈路即可

代码

#include <iostream>
#include <cstring>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <cstdio>
#include <cstdlib>
#include <climits>
#include <deque>
#include <bitset>
#include <algorithm>
using namespace std;

#define PI acos(-1.0)
#define LL long long
#define PII pair<int, int>
#define PLL pair<LL, LL>
#define mp make_pair
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "wb", stdout)
#define scan(x) scanf("%d", &x)
#define scan2(x, y) scanf("%d%d", &x, &y)
#define scan3(x, y, z) scanf("%d%d%d", &x, &y, &z)
#define sqr(x) (x) * (x)
#define pr(x) cout << #x << " = " << x << endl
#define lc o << 1
#define rc o << 1 | 1
#define pl() cout << endl

const int maxn = 10005;
int n, m, p[maxn], r[maxn];
bool use[maxn];
vector<PII> G[105];
struct Edge {
    int u, v, w;
    Edge(int a, int b, int c) : u(a), v(b), w(c) {

    }
    Edge() {

    }
} E[maxn];

int find(int x) {
    return p[x] == x ? x : p[x] = find(p[x]);
}

bool cmp(Edge x, Edge y) {
    return x.w < y.w;
}

int kruskal() {
    for (int i = 1; i <= n; i++) p[i] = i;
    int ans = 0;
    sort(E, E + m, cmp);
    for (int i = 0; i < m; i++) {
        int x = E[i].u, y = E[i].v;
        x = find(x);
        y = find(y);
        if (x != y) {
            ans += E[i].w;
            p[x] = y;
            use[i] = 1;
            G[E[i].u].push_back(mp(E[i].v, E[i].w));
            G[E[i].v].push_back(mp(E[i].u, E[i].w));
        }
    }
    return ans;
}

const int maxnode = 105;
int f[maxnode][maxnode], x;
void dfs(int u, int v, int MAX, int w) {
    if (v > 0) f[x][u] = max(MAX, w);
    for (int i = 0; i < G[u].size(); i++) {
        int son = G[u][i].first;
        if (son == v) continue;
        dfs(son, u, f[x][u], G[u][i].second);
    }
}

int main() {
    int T;
    scan(T);
    while (T--) {
        scan2(n, m);
        memset(f, 0, sizeof(f));
        memset(use, 0, sizeof(use));
        for (int i = 0; i < 105; i++) G[i].clear();
        for (int i = 0; i < m; i++) {
            int x, y, z;
            scan3(E[i].u, E[i].v, E[i].w);
        }
        int mst = kruskal();
        for (int i = 1; i <= n; i++) {
            x = i;
            dfs(i, -1, 0, 0);
        }
        bool flag = true;
        for (int i = 0; i < m; i++) {
            if (use[i]) continue;
            int from = E[i].u, to = E[i].v;
            int mst_ans = max(f[from][to], f[to][from]);
            if (mst_ans == E[i].w) {
                flag = false;
                break;
            }
        }
        if (!flag) puts("Not Unique!");
        else printf("%d\n", mst);
    }
    return 0;
}