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已解决
王子健
初级天翼
初级天翼
# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template <class Num> inline void Cmax(Num &x, const Num y) {
x = y > x ? y : x;
}
template <class Num> inline void Cmin(Num &x, const Num y) {
x = y < x ? y : x;
}
const int maxn(5005);
const int maxm(2e5 + 5);
const double eps(1e-8);
int n, m, first[maxn], cnt, vis[maxn], rt[maxn], tot, cov[maxm << 1], ans, fa[maxn];
double se, e, dis[maxn];
priority_queue < pair <double, int> > q;
struct Heap {
int ls, rs, dis, ed;
double w;
} tr[maxm * 20];
struct Edge {
int to, next;
double w;
} edge[maxm << 1];
inline void Add(int u, int v, double w) {
edge[cnt] = (Edge){v, first[u], w}, first[u] = cnt++;
edge[cnt] = (Edge){u, first[v], w}, first[v] = cnt++;
}
inline int NewNode(double w, int ed) {
int x = ++tot;
tr[x].w = w, tr[x].dis = 1, tr[x].ed = ed;
return x;
}
int Merge(int x, int y) {
if (!x || !y) return x + y;
if (tr[x].w - tr[y].w >= eps) swap(x, y);
int p = ++tot;
tr[p] = tr[x], tr[p].rs = Merge(tr[p].rs, y);
if (tr[tr[p].ls].dis < tr[tr[p].rs].dis) swap(tr[p].ls, tr[p].rs);
tr[p].dis = tr[tr[x].rs].dis + 1;
return p;
}
void Dfs(int u) {
vis[u] = 1;
for (int e = first[u], v; e != -1; e = edge[e].next)
if (e & 1) {
double w = edge[e].w;
if (fabs(dis[u] + w - dis[v = edge[e].to]) < eps && !vis[v])
fa[v] = u, cov[e ^ 1] = 1, Dfs(v);
}
}
int main() {
memset(first, -1, sizeof(first));
memset(dis, 127, sizeof(dis));
scanf("%d%d%lf", &n, &m, &se);
for (int i = 1, u, v; i <= m; ++i) scanf("%d%d%lf", &u, &v, &e), Add(u, v, e);
dis[n] = 0, q.push(make_pair(0, n));
while (!q.empty()) {
int u = q.top().second;
q.pop();
if (vis[u]) continue;
vis[u] = 1;
for (int e = first[u]; ~e; e = edge[e].next)
if (e & 1) {
int v = edge[e].to;
if (dis[v] - (dis[u] + edge[e].w) >= eps)
q.push(make_pair(-(dis[v] = dis[u] + edge[e].w), v));
}
}
for (int i = 1; i <= n; ++i) vis[i] = 0;
Dfs(n);
for (int e = 0, u, v; e < cnt; e += 2)
if (!cov[e]) {
u = edge[e ^ 1].to, v = edge[e].to;
if (dis[u] == dis[0] || dis[v] == dis[0]) continue;
rt[u] = Merge(rt[u], NewNode(dis[v] + edge[e].w - dis[u], v));
}
for (int i = 1; i <= n; ++i) q.push(make_pair(-dis[i], i));
for (int i = 1, u; i <= n; ++i) {
u = q.top().second, q.pop();
if (fa[u]) rt[u] = Merge(rt[u], rt[fa[u]]);
}
if (dis[1] - se < eps) se -= dis[1], ++ans;
if (rt[1]) q.push(make_pair(-tr[rt[1]].w, rt[1]));
while (!q.empty()) {
int ed = q.top().second;
double cur = q.top().first, w = dis[1] - cur;
if (w - se >= eps) break;
q.pop(), se -= w, ++ans;
for (int i = 0; i < 2; ++i) {
int nxt = i ? tr[ed].rs : tr[ed].ls;
if (nxt) q.push(make_pair(cur + tr[ed].w - tr[nxt].w, nxt));
}
if (rt[tr[ed].ed]) q.push(make_pair(cur - tr[rt[tr[ed].ed]].w, rt[tr[ed].ed]));
}
printf("%d\n", ans);
return 0;
}
洛谷上的题,93分,望解答。