CodeForces - 25D Roads not only in Berland (并查集)

分析：利用并查集处理每一条边上两点的关系。假设两个点已直接或间接可互达，在此之外又连有额外的边，那么该条边应该被重建（集合合并问题）。

#include
using namespace std;
const int N = 1e3 + 5;
typedef pair P;
int fa[N], Rank[N];
int find(int x) {
    if(x == fa[x])
        return x;
    else
        return fa[x] = find(fa[x]);
}
bool Merge(int u, int v) {
    if(u > v)
        swap(u, v);
    int x = find(u);
    int y = find(v);
    if(x != y) {
        fa[y] = x;
        Rank[x] += Rank[y];
        return true;
    }
    return false;
}
struct data {
    int i, j, u, v;
    data(int i, int j, int u, int v): i(i), j(j), u(u), v(v) { }
};
int main() {
    int n, i, j;
    scanf("%d", &n);
    for(i = 1; i <= n; i += 1) {
        fa[i] = i;
        Rank[i] = 1;
    }
    int x, y;
    vector
 v;
    for(i = 1; i < n; i += 1) {
        scanf("%d%d", &x, &y);
        if(!Merge(x, y)){
            v.push_back(P(x, y)); //应该被重建的边
        }
    }
    if(Rank[find(n)] == n)
        printf("0\n");
    else {
        vector Set;
        for(i = 1; i <= n; i += 1) {
            if(find(i) == i){ //处理出目前所有的集合
                Set.push_back(i);
            }
        }
        bool vis[N];
        memset(vis, false, sizeof(vis));
        vector res;
        for(i = 0; i < Set.size(); i += 1) {
            for(j = 0; j < v.size(); j += 1) {
                if(vis[j]){
                    continue;
                }
                if(Merge(v[j].first, Set[i])) { //集合合并
                    vis[j] = true;
                    res.push_back(data(v[j].first, v[j].second, v[j].first, Set[i]));
                }
            }
        }
        printf("%d\n", (int)res.size());
        for(i = 0; i < (int)res.size(); i += 1)
            printf("%d %d %d %d\n", res[i].i, res[i].j, res[i].u, res[i].v);
    }
    return 0;
}
/*
7
1 2
1 3
2 3
3 4
4 5
3 5
*/