Codeforces 1076D Edge Deletion——最短路+dfs

题意：

n个点m条边的无向连通图（无自环、重边），现在要删除一些边使得图最多剩下k条边，并且这样的点尽量多：这个点到1号节点的最短路不变

思路：

建出最短路树，统计边数，然后跑一遍dfs，在回溯的时候判断边数与k的关系，若边数>k就使边数减一，否则记录答案

#include 
using namespace std;
const int maxn = 3e5 + 10;
typedef long long ll;
const ll INF = 1e16;
typedef pair P;
int n, m, k;
int tot, head[maxn];
struct Edge {
    int to, cost, id, next;
}edges[2*maxn];
void init_edges() {
    tot = 0;
    memset(head, -1, sizeof(head));
}
void addedge(int from, int to, int cost, int id) {
    edges[tot].to = to; edges[tot].cost = cost; edges[tot].id = id;
    edges[tot].next = head[from]; head[from] = tot++;
}
bool vis[maxn];
ll dis[maxn];
struct Path {
    int to;
    int id;
}path[maxn];
void dijkstra() {
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < maxn; i++) dis[i] = INF;
    memset(path, 0, sizeof(path));
    int s = 1;
    dis[s] = 0;
    priority_queue, greater
 > que;
    que.push(make_pair(0, s));
    while (!que.empty()) {
        int u = que.top().second; que.pop();
        if (vis[u]) continue;
        vis[u] = 1;
        for (int i = head[u]; ~i; i = edges[i].next) {
            int v = edges[i].to, cost = edges[i].cost, id = edges[i].id;
            if (dis[v] > dis[u] + cost) {
                dis[v] = dis[u] + cost;
                path[v].to = u;
                path[v].id = id;
                que.push(make_pair(dis[v], v));
            }
        }
    }
}
int cnt;
void dfs(int u) {
    if (u == 0) return;
    if (vis[u]) return;
    vis[u] = 1;
    int v = path[u].to, id = path[u].id;
    if (v == 0) return;
    addedge(v, u, 1, id);
    cnt++;
    dfs(v);
}
vector ans;
void solve(int u) {
    for (int i = head[u]; ~i; i = edges[i].next) {
        int v = edges[i].to, id = edges[i].id;
        solve(v);
        if (cnt > k) {
            cnt--;
        }
        else {
            ans.push_back(id);
        }
    }
}
int main() {
    scanf("%d%d%d", &n, &m, &k);
    init_edges();
    for (int i = 1; i <= m; i++) {
        int u, v, cost;
        scanf("%d%d%d", &u, &v, &cost);
        addedge(u, v, cost, i);
        addedge(v, u, cost, i);
    }
    dijkstra();
    cnt = 0;
    init_edges();
    memset(vis, 0, sizeof(vis));
    for (int i = 2; i <= n; i++) {
        dfs(i);
    }
    ans.clear();
    solve(1);
    sort(ans.begin(), ans.end());
    printf("%d\n", ans.size());
    for (int i = 0; i < ans.size(); i++) {
        printf("%d", ans[i]);
        if (i == ans.size()-1) printf("\n");
        else printf(" ");
    }
    return 0;
}