传送门：https://www.luogu.org/problemnew/show/P1027
题意：
图中有n个城市，每个城市有4个机场在矩形的四个顶点上。一个城市间的机场可以通过高铁通达，不同城市间要通过飞机。现在问从s到t城市最少需要多少的费用。
思路：
已知矩形的三个顶点，可以用勾股定理确定斜边后，利用平行四边形原理——两对对角顶点的x之和是相同的，y之和也是相同的得到第四个顶点。然后用求最短路的dji即可。
#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#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 pair<double,int>pdi;
//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; //
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
ll mod = ;
const double esp = 1e-;
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=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
} /*-----------------------showtime----------------------*/
const int maxn = ;
int c1,s,t,n;
double dis[maxn];
struct node
{
int x,y,bl;
int cst;
}a[maxn];
void getp(int x1,int y1,int x2,int y2,int x3,int y3,int i){
int ab = (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2);
int ac = (x1-x3)*(x1-x3) + (y1-y3)*(y1-y3);
int bc = (x2-x3)*(x2-x3) + (y2-y3)*(y2-y3);
int x4,y4; if(ab + ac == bc) x4 = x2 + x3 - x1, y4 = y2 + y3 - y1;
if(ab + bc == ac) x4 = x1 + x3 - x2, y4 = y1 + y3 - y2;
if(ac + bc == ab) x4 = x1 + x2 - x3, y4 = y1 + y2 - y3;
a[i+].x = x4;
a[i+].y = y4;
}
double getdis(int i,int j){
return sqrt(1.0*(a[i].x - a[j].x)*(a[i].x - a[j].x) + 1.0*(a[i].y - a[j].y)*(a[i].y - a[j].y));
}
void dji(){
for(int i=; i<=*n; i++)dis[i] = 1000000000.9;
dis[(s-) * +] = dis[(s-) * +] = dis[(s-) * +] = dis[(s-) * + ] =;
priority_queue<pdi>que;
que.push(pdi(0.0,(s-) * +));
que.push(pdi(0.0,(s-) * +));
que.push(pdi(0.0,(s-) * +));
que.push(pdi(0.0,(s-) * +));
while(!que.empty()){
pdi tmp = que.top(); que.pop();
if(dis[tmp.se] < -*tmp.fi)continue;
for(int i=; i<=*n; i++){
if(tmp.se != i){
double d = getdis(tmp.se, i);
if(a[tmp.se].bl == a[i].bl) {
if(dis[i] > dis[tmp.se] +1.0* a[i].cst * d){
dis[i] = dis[tmp.se] + 1.0*a[i].cst * d;
que.push(pdi(-dis[i],i));
}
}
else {
if(dis[i] > dis[tmp.se] + 1.0*c1 * d){
dis[i] = dis[tmp.se] + 1.0*c1 * d;
que.push(pdi(-dis[i],i));
}
}
}
}
} }
int main(){
int T; scanf("%d", &T);
while(T--){
scanf("%d%d%d%d",&n, &c1, &s, &t);
for(int i=; i<=*n; i+=){
int cst;
scanf("%d%d%d%d%d%d%d", &a[i].x,&a[i].y,&a[i+].x, &a[i+].y,&a[i+].x, &a[i+].y,&cst);
getp(a[i].x, a[i].y,a[i+].x, a[i+].y,a[i+].x, a[i+].y,i);
a[i+].cst = a[i].cst = a[i+].cst = a[i+].cst = cst;
a[i+].bl = a[i].bl = a[i+].bl = a[i+].bl = (i-)/ + ;
} dji();
double ans = 1000000000.9;
ans = min(ans, dis[(t-)*+]);
ans = min(ans, dis[(t-)*+]);
ans = min(ans, dis[(t-)*+]);
ans = min(ans, dis[(t-)*+]);
printf("%.1f\n", ans); }
return ;
}
P1027