Edge Deletion--最短路+bfs

題目鏈接：https://vjudge.net/problem/CodeForces-1076D

題目大意：給出n個點，m條邊，最多能保留K條邊。原圖上每個點的最短路爲di。現在刪除了邊後，使得到點1的距離仍爲di的點數量最多的情況下，輸出應該保留的K條邊的編號。

分析：既然要使結果最大，那就儘可能多保留邊，即k條邊。跑一遍最短路，求出1到每個點的最短距離，然後bfs，如果某條邊在最短路的路徑上，那麼輸出這條邊。注意用long long。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
#define inf 0x3f3f3f3f
typedef long long ll;
const int N = 3e5 + 5;
int n, m, k, cnt, head[2 * N];
ll dis[N];
bool vis[N];
vector<int> ans;
struct Edge {
    int id, to, nxt;
    ll w;
}edge[2 * N];
struct node {
    int id;
    ll d;
    node(){}
    node(int id_, ll d_) {
        this->id = id_;
        this->d = d_;
    }
    bool operator < (const node &a) const {
        return a.d < d;
    }
};
void init() {
    cnt = 0;
    memset(head, -1, sizeof head);
}
void add(int id, int a, int b, ll w) {
    edge[cnt].id = id;
    edge[cnt].to = b;
    edge[cnt].w = w;
    edge[cnt].nxt = head[a];
    head[a] = cnt++;
}
void dijkstra() {
    memset(dis, inf, sizeof dis);
    dis[1] = 0;
    priority_queue<node> q;
    q.push(node(1, 0));
    node tmp;
    while(!q.empty()) {
        tmp = q.top();
        q.pop();
        if(vis[tmp.id]) continue;
        vis[tmp.id] = 1;
        for(int i = head[tmp.id]; i != -1; i = edge[i].nxt) {
            int j = edge[i].to;
            if(!vis[j] && dis[j] > dis[tmp.id] + edge[i].w) {
                dis[j] = dis[tmp.id] + edge[i].w;
                q.push(node(j, dis[j]));
            }
        }
    }
}
void bfs() {
    memset(vis, 0, sizeof vis);
    vis[1] = 1;
    queue<int> q;
    q.push(1);
    int x;
    while(!q.empty()) {
        x = q.front();
        q.pop();
        for(int i = head[x]; i != -1; i = edge[i].nxt) {
            int j = edge[i].to;
            if(!vis[j] && dis[j] == dis[x] + edge[i].w) {
                vis[j] = 1;
                ans.push_back(edge[i].id);
                q.push(j);
            }
        }
    }
}
void print() {
    int cnt = ans.size();
    cnt = min(cnt, k);
    printf("%d\n", cnt);
    for(int i = 0; i < cnt; i++)
        printf("%d%s", ans[i], i == cnt - 1 ? "\n" : " ");
}
int main() {
    init();
    scanf("%d %d %d", &n, &m, &k);
    int a, b;
    ll w;
    for(int i = 1; i <= m; i++) {
        scanf("%d %d %lld", &a, &b, &w);
        add(i, a, b, w);
        add(i, b, a, w);
    }
    dijkstra();
    bfs();
    print();
    return 0;
}