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; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 ll 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 = 500; 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+3].x = x4; a[i+3].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=1; i<=4*n; i++)dis[i] = 1000000000.9; dis[(s-1) * 4+1] = dis[(s-1) * 4+2] = dis[(s-1) * 4+3] = dis[(s-1) * 4 + 4] =0; priority_queue<pdi>que; que.push(pdi(0.0,(s-1) * 4+1)); que.push(pdi(0.0,(s-1) * 4+2)); que.push(pdi(0.0,(s-1) * 4+3)); que.push(pdi(0.0,(s-1) * 4+4)); while(!que.empty()){ pdi tmp = que.top(); que.pop(); if(dis[tmp.se] < -1*tmp.fi)continue; for(int i=1; i<=4*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=1; i<=4*n; i+=4){ int cst; scanf("%d%d%d%d%d%d%d", &a[i].x,&a[i].y,&a[i+1].x, &a[i+1].y,&a[i+2].x, &a[i+2].y,&cst); getp(a[i].x, a[i].y,a[i+1].x, a[i+1].y,a[i+2].x, a[i+2].y,i); a[i+3].cst = a[i].cst = a[i+1].cst = a[i+2].cst = cst; a[i+3].bl = a[i].bl = a[i+1].bl = a[i+2].bl = (i-1)/4 + 1; } dji(); double ans = 1000000000.9; ans = min(ans, dis[(t-1)*4+1]); ans = min(ans, dis[(t-1)*4+2]); ans = min(ans, dis[(t-1)*4+3]); ans = min(ans, dis[(t-1)*4+4]); printf("%.1f\n", ans); } return 0; }