文章图片
题意 【Codeforces ~ 1076D ~ Edge Deletion （最短路，堆优化理解）】给你一个n个点，m条边的DAG图，边为双向边，没有重边。现在最多保留k条边，怎么使得好点个数最多。
好点定义为：在原图中1到该点距离和只保留某一些边后的图中1到该点距离不变的点。
先输出保留边的个数，然后输出这些保留的边的编号（1~m）。
思路 堆优化dijkstra中，更新前 k 个点用到的边就是答案。
数据比较大，注意开long long
#include
using namespace std;
const int MAXN = 3e5+5;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
struct Edge
{
int from, to;
LL dist;
//起点,终点,距离
Edge(int from, int to, LL dist):from(from), to(to), dist(dist) {}
};
struct Dijkstra
{
int n, m;
//结点数,边数(包括反向弧)
vector edges;
//边表。edges[e]和edges[e^1]互为反向弧
vector G[MAXN];
//邻接表，G[i][j]表示结点i的第j条边在edges数组中的序号
int vis[MAXN];
//标记数组
LL d[MAXN];
//s到各个点的最短路
int p[MAXN];
//上一条弧void init(int n)
{
this->n = n;
edges.clear();
for (int i = 0;
i <= n;
i++) G[i].clear();
}void AddEdge(int from, int to, int dist)
{
edges.emplace_back(from, to, dist);
m = edges.size();
G[from].push_back(m - 1);
}struct HeapNode
{
int from;
LL dist;
bool operator < (const HeapNode& rhs) const
{
return rhs.dist < dist;
}
HeapNode(int u, LL w): from(u), dist(w) {}
};
void dijkstra(int s, vector& ans, int k)
{
priority_queue Q;
for (int i = 0;
i <= n;
i++) d[i] = INF;
memset(vis, 0, sizeof(vis));
d[s] = 0;
Q.push(HeapNode(s, 0));
while (!Q.empty() && ans.size() <= k)
{
HeapNode x = Q.top();
Q.pop();
int u = x.from;
if (vis[u]) continue;
vis[u] = true;
ans.push_back(p[u]/2);
for (int i = 0;
i < G[u].size();
i++)
{
Edge& e = edges[G[u][i]];
if (d[e.to] > d[u] + e.dist)
{
d[e.to] = d[u] + e.dist;
p[e.to] = G[u][i];
Q.push(HeapNode(e.to, d[e.to]));
}
}
}
}
}gao;
int n, m, k;
int vis[MAXN];
int main()
{
scanf("%d%d%d", &n, &m, &k);
memset(vis, 0, sizeof(vis));
gao.init(n);
while (m--)
{
int x, y, w;
scanf("%d%d%d", &x, &y, &w);
gao.AddEdge(x, y, w);
gao.AddEdge(y, x, w);
}
vector ans;
gao.dijkstra(1, ans, k);
printf("%d\n", ans.size()-1);
for (int i = 1;
i < ans.size();
i++)
printf("%d%c", ans[i]+1, i==ans.size()-1?'\n':' ');
return 0;
}
