牛客2018国庆集训派对Day3 I Metropolis 多源最短路径

传送门：https://www.nowcoder.com/acm/contest/203/I

 

题意：

　　求每个大都会到最近的一个大都会的距离。

思路：

　　把每个大都会设为起点，跑一遍最短路。在跑最短路的时候，假设有一个大都会A，它不能更新一个非大都会B点，就说明这个B到另一个大都会C很近，用$dis[u] + dis[B] + w[u][B]$更新A点的答案。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include   <complex>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>
 
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)
 
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue
 
 
 
typedef long long ll;
typedef unsigned long long ull;
 
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;
typedef complex<double> cp;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'
 
#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;
 
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
// const int mod = 10007;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;
 
 
template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}
 
 
/*-----------------------showtime----------------------*/
            const int maxn = 2e5+9;
            ll ans[maxn],dis[maxn];
            int star[maxn],from[maxn];
            int n,m,q;
            vector<pll>mp[maxn];
            void dji(){
                priority_queue<pll>que;
                for(int i=1; i<=n; i++)dis[i] = inff;
                for(int i=1; i<=q; i++)dis[star[i]] = 0,que.push(pll(0,star[i]));

                while(!que.empty()){
                    pll f = que.top();que.pop();
                    if(dis[f.se] < -1ll*f.fi)continue;
                    int u = f.se;
                    for(int i=0; i<mp[u].size(); i++){
                        int v = mp[u][i].fi;
                        if(dis[v] > dis[u] + mp[u][i].se){
                            dis[v] = dis[u] + mp[u][i].se;
                            from[v] = from[u];
                            que.push(pll(-1ll*dis[v],v));
                        }
                        else if(from[v] != from[u]){
                            ans[from[u]] = min(ans[from[u]] , dis[u] + dis[v] + mp[u][i].se);
                            ans[from[v]] = min(ans[from[v]] , dis[u] + dis[v] + mp[u][i].se);
                        }
                    }

                }

            }
int main(){
            
            scanf("%d%d%d", &n, &m, &q);
            for(int i=1; i<=q; i++)scanf("%d",&star[i]),from[star[i]] = star[i];
            for(int i=1; i<=n; i++)ans[i] = inff;
            for(int i=1; i<=m; i++){
                int u,v,w;
                scanf("%d%d%d", &u, &v, &w);
                mp[u].pb(pll(v,w));
                mp[v].pb(pll(u,w));
            }
            dji();
            for(int i=1; i<=q; i++){
                if(i==q)printf("%lld\n", ans[star[i]]);
                else printf("%lld ",ans[star[i]]);
            }
            return 0;
}