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];
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 = SIZ[u];
    for(vector<int>::iterator v = G[u].begin(); v != G[u].end(); ++v){
//      if(*v == fa) continue;
        siz += DFS(*v, siz);
    }
    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, 0);
    
    printf("%lld", ans);
    
    return 0;
} 

原文地址：https://www.cnblogs.com/bingoyes/p/11745522.html