最短路径（巧妙的矩阵交换）

http://poj.org/problem?id=3268

题意：N头牛分别在N个农场有M条边的无向边，问所有牛中前往X农场，并返回走的最长距离的牛的距离。

解法：先求以X为源点到其他各农场的最短距离（相当于各牛返回），然后将M边反向，再以X为源点求到各农场的距离（相当于各牛去往X），统计每头牛走的距离，输出最大距离。

 

//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
#include <stdlib.h>
using namespace std;
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd(m,n) __gcd(m, n)
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
int lcm(int a , int b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define size(v) (int)(v.size())
#define cin(x) scanf("%lld" , &x);
const int N = 1e7+9;
const int maxn = 1e5+9;
const double esp = 1e-6;
int head[maxn] , tol ;
int n , m , x ;
struct node{
    int to , w;
    bool operator < (const node x) const{
        return w > x.w ;
    }
    node(int a , int b){to = a , w = b;}
    node(){}
};
struct Graph{
    int to , w , next;
}g[maxn<<1];

void add(int u , int v , int w){
    g[++tol] = {v , w , head[u]};
    head[u] = tol;
}
int vis[maxn] , dis[maxn] ,ans[maxn];
void dijia(int r){
    ME(vis , 0);
    rep(i , 1 , n)dis[i] = INF; dis[r] = 0 ;
    priority_queue<node>q;
    q.push(node(r , dis[r]));
    node now;
    while(!q.empty()){
        now = q.top();q.pop();
        if(vis[now.to]) continue;
        for(int i = head[now.to] ; i ; i = g[i].next){
            int v = g[i].to;
            if(dis[v] > dis[now.to] + g[i].w){
                dis[v] = dis[now.to] + g[i].w;
                q.push(node(v , dis[v]));
            }
        }
    }
}
void init(){
    tol = 0 ;
    ME(head , 0);
}
int u[maxn] , v[maxn], w[maxn];
void solve(){
    scanf("%lld%lld%lld" , &n , &m , &x);
    rep(i , 1 , m){
        scanf("%lld%lld%lld" , &u[i] , &v[i] , &w[i]);
        add(u[i] , v[i] , w[i]);
    }
    dijia(x);
    rep(i , 1 , n) ans[i] += dis[i];
    init();
    rep(i , 1 , m){
        add(v[i] , u[i] , w[i]);
    }
    dijia(x);
    rep(i , 1 , n) ans[i] += dis[i];
    int ma = -INF;
    rep(i , 1 , n) ma = max(ma , ans[i]);
    cout << ma << endl;
}

signed main()
{
    //int t;
    //scanf("%lld" , &t);
    //while(t--)
        solve();
}