LuoguP5201 [USACO19JAN]Shortcut(最短路树)
字典序?建树时从小枚举,用\(Dijkstra\)的血泪建好树,\(size\)大小决定贡献
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); a <= (c); ++a)
#define nR(a,b,c) for(register int a = (b); a >= (c); --a)
#define Fill(a,b) memset(a, b, sizeof(a))
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))
#define QWQ
#ifdef QWQ
#define D_e_Line printf("\n---------------\n")
#define D_e(x) cout << (#x) << " : " << x << "\n"
#define Pause() system("pause")
#define FileOpen() freopen("in.txt", "r", stdin)
#define FileSave() freopen("out.txt", "w", stdout)
#define TIME() fprintf(stderr, "\nTIME : %.3lfms\n", clock() * 1000.0 / CLOCKS_PER_SEC)
#else
#define D_e_Line ;
#define D_e(x) ;
#define Pause() ;
#define FileOpen() ;
#define FileSave() ;
#define TIME() ;
#endif
struct ios {
template<typename ATP> inline ios& operator >> (ATP &x) {
x = 0; int f = 1; char c;
for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
while(c >= '0' && c <='9') x = x * 10 + (c ^ '0'), c = getchar();
x *= f;
return *this;
}
}io;
using namespace std;
template<typename ATP> inline ATP Max(ATP a, ATP b) {
return a > b ? a : b;
}
template<typename ATP> inline ATP Min(ATP a, ATP b) {
return a < b ? a : b;
}
template<typename ATP> inline ATP Abs(ATP a) {
return a < 0 ? -a : a;
}
#include <queue>
#include <vector>
const int N = 1e4 + 7;
const int M = 5e4 + 7;
#define int long long
int n, m, T;
int SIZ[N], dis[N];
struct Edge {
int nxt, pre, w;
} e[M << 1];
int head[N], cntEdge;
inline void add(int u, int v, int w) {
e[++cntEdge] = (Edge){ head[u], v, w}, head[u] = cntEdge;
}
struct nod {
int x, w;
bool operator < (const nod &com) const {
return w > com.w;
}
};
priority_queue<nod> q;
inline void Dijkstra(int st) {
Fill(dis, 0x3f);
dis[st] = 0;
q.push((nod){ st, 0});
while(!q.empty()){
int u = q.top().x, w = q.top().w;
q.pop();
if(w != dis[u]) continue;
for(register int i = head[u]; i; i = e[i].nxt){
int v = e[i].pre;
if(dis[v] > dis[u] + e[i].w){
dis[v] = dis[u] + e[i].w;
q.push((nod){ v, dis[v]});
}
}
}
}
int ans;
vector<int> G[N];
bool vis[N];
inline void Build() {
R(u,1,n){
for(register int i = head[u]; i; i = e[i].nxt){
int v = e[i].pre;
if(dis[v] == dis[u] + e[i].w && !vis[v]){
vis[v] = 1;
G[u].push_back(v);
}
}
}
}
inline int DFS(int u) {
int siz = SIZ[u];
for(vector<int>::iterator v = G[u].begin(); v != G[u].end(); ++v){
// if(*v == fa) continue;
siz += DFS(*v);
}
ans = Max(ans, siz * (dis[u] - T));
return siz;
}
#undef int
int main() {
#define int long long
// FileOpen();
io >> n >> m >> T;
R(i,1,n){
io >> SIZ[i];
}
R(i,1,m){
int u, v, w;
io >> u >> v >> w;
add(u, v, w);
add(v, u, w);
}
Dijkstra(1);
Build();
DFS(1);
printf("%lld", ans);
return 0;
}