noip水题系列
https://www.luogu.org/problem/P1027
题目明显就是一个多组数据(也就<=10)单源最短路,
这里可以就不用floyed
但你会发现它是道蓝题也是有一定道理的
你会很恼火它的建边:
使用勾股定理加一系列的特判,而且又是浮点数
话不多述,
思路简单,代码麻烦,就把题解修改一下搬过来了
code:
#include
#include
#include
#include
using namespace std;
struct data {
int x,y;
int city;
};
const int maxn=100;
int s,t,A,B;
int T[maxn+1];
double dis[maxn<<2|1];
data a[maxn<<2|1];
int pingfang(int x) { return x*x; }
double juli(int x1, int y1, int x2, int y2) { return sqrt(pingfang(x1-y1)+pingfang(x2-y2)); }
void get_4th(int x1, int y1, int x2, int y2, int x3, int y3, int i) {
int ab=pingfang(x1-x2)+pingfang(y1-y2),
ac=pingfang(x1-x3)+pingfang(y1-y3),
bc=pingfang(x2-x3)+pingfang(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;
}
void init() {
memset(a,0,sizeof(a));
scanf("%d%d%d%d",&s,&t,&A,&B);
for (int i=1; i<=4*s; i+=4) {
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,&T[i/4+1]);
a[i].city=a[i+1].city=a[i+2].city=a[i+3].city=i/4+1;
get_4th(a[i].x,a[i].y,a[i+1].x,a[i+1].y,a[i+2].x,a[i+2].y,i);
}
}
void spfa() {
bool mark[maxn<<2|1];
queue q;
for (int i=1; i<=4*s; i++) dis[i]=99999999.99999;
for (int i=A*4-3;i<=A*4;i++)
dis[i]=0, q.push(i), mark[i]=true;
while (!q.empty()) {
int x=q.front(); q.pop(); mark[x]=false;
for (int i=1; i<=4*s; i++) {
if (i==x) continue;
double cost=juli(a[x].x,a[i].x,a[x].y,a[i].y);
if (a[i].city==a[x].city) cost*=T[a[i].city];
else cost*=t;
if (dis[x]+cost