CodeForces - 240E Road Repairs(最小树形图＋输出路径)

题目大意：有N个城市，城市1是首都
有M条路，路的状态有两种，一种是可行的，另一种是不可行的，不可行的可以修成可行，但要花1单位的费用
现在问，要让首都能到任意城市，需要花费的最小代价是多少，需要修的路是哪几条

解题思路：就是最小树形图＋输出路径了

#include <cstdio>
#include <cstring>
using namespace std;
const int N = 2000010;
const int INF = 0x3f3f3f3f;

struct Edge{
    int u, v, d, dd, id;
    Edge() {}
    Edge(int u, int v, int d, int id): u(u), v(v), d(d), id(id), dd(d){};
};

const int M = 2000010;
struct Used{
    int pre, id;
}U[M];
int UseEdge[M];

struct DirectorMT{
    int n, m, tot;
    Edge edges[N];
    int vis[N], id[N], in[N], pre[N];
    int preEdge[N];

    void init(int n) {
        this->n = n;
        m = 0;
    }

    void AddEdge(int u, int v, int mark, int id) {
        edges[m++] = Edge(u, v, mark, id);
    }

    int DirMt(int root) {

        tot = m;
        int ans = 0;
        while (1) {
            for (int i = 0; i < n; i++) in[i] = INF;

            for (int i = 0; i < m; i++) {
                int u = edges[i].u;
                int v = edges[i].v;

                if (edges[i].d < in[v] && u != v ) {
                    in[v] = edges[i].d;
                    pre[v] = u;
                    preEdge[v] = edges[i].id;
                }
            }

            for (int i = 0; i < n; i++) {
                if (i == root) continue;
                if (in[i] == INF) return -1;
            }

            memset(id, -1, sizeof(id));
            memset(vis, -1, sizeof(vis));
            in[root] = 0;
            int cnt = 0;
            for (int i = 0; i < n; i++) {
                ans += in[i];
                int v = i;
                //该边被使用了
                if (i != root) UseEdge[preEdge[i]]++;
                while (vis[v] != i && v != root && id[v] == -1) {
                    vis[v] = i;
                    v = pre[v];
                }

                if (v != root && id[v] == -1) {
                    for (int u = pre[v]; u != v; u = pre[u])
                        id[u] =  cnt;
                    id[v] = cnt++;
                }
            }

            if (cnt == 0) break;
            for (int i = 0; i < n; i++)
                if (id[i] == -1) id[i] = cnt++;

            for (int i = 0; i < m; i++) {
                int v = edges[i].v;
                edges[i].v = id[edges[i].v];
                edges[i].u = id[edges[i].u];
                //因为有新的边进来了，如果用到该边的话，那么上一条边就要被取消掉了
                if (edges[i].u != edges[i].v) {
                    edges[i].d -= in[v];
                    U[tot].id = edges[i].id;
                    U[tot].pre = preEdge[v];
                    edges[i].id = tot;
                    tot++;
                }
            }
            n = cnt;
            root = id[root];
        }

        for (int i = tot - 1; i >= m; i--)
            if (UseEdge[i]) {
                UseEdge[U[i].id]++; 
                UseEdge[U[i].pre]--;
            }
        return ans;

    }
}MT;

int n, m;
void init() {
    MT.init(n);
    int u, v, mark;
    for (int i = 0; i < m; i++) {
        scanf("%d%d%d", &u, &v, &mark);
        u--; v--;
        MT.AddEdge(u, v, mark, i);
    }
    int ans = MT.DirMt(0);

    if (ans == -1 || ans == 0) { 
        printf("%d\n", ans);
        return ;
    }
    printf("%d\n", ans);
    for (int i = 0; i < m; i++)
        if (UseEdge[i] && MT.edges[i].dd) 
            printf("%d\n", i + 1);
}

int main() {

    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    scanf("%d%d", &n, &m);
    init();
    return 0;

}