Pku2457 Part Acquisition problem1653

蒟蒻是第一次写题解，不喜勿喷，好了，先看题目：

The cows have been sent on a mission through space to acquire a new milking machine for their barn. They are flying through a cluster of stars containing N (1 <= N <= 50,000) planets, each with a trading post.
The cows have determined which of K (1 <= K <= 1,000) types of objects (numbered 1…K) each planet in the cluster desires, and which products they have to trade. No planet has developed currency, so they work under the barter system: all trades consist of each party trading exactly one object (presumably of different types).
The cows start from Earth with a canister of high quality hay (item 1), and they desire a new milking machine (item K). Help them find the best way to make a series of trades at the planets in the cluster to get item K. If this task is impossible, output -1.
给出N条边，单向的。再给出数字K.问从1号点到K点号，最短路上经过多少个点。
Input

• Line 1: Two space-separated integers, N and K.
• Lines 2…N+1: Line i+1 contains two space-separated integers, a_i and b_i respectively, that are planet i’s trading trading products. The planet will give item b_i in order to receive item a_i.
  Output
• Line 1: One more than the minimum number of trades to get the milking machine which is item K (or -1 if the cows cannot obtain item K).
  Sample Input
  6 5
  1 3
  3 2
  2 3
  3 1
  2 5
  5 4
  Sample Output
  4
  ————————————————————————————————————
  题目大意就是给你n条路，边权都为1，然后n行描述哪条连哪条，然后再输出1到k的最短路经过点的个数。
  这道题的话我们就用dijkstra最短路（具体百度），而不用floyed最短路，虽然不能解释具体原因，但至少时间复杂度上要快很多，所以说到这里各位神犇想必一定有思路了，就是读入后一遍dijkstra最短路，然后点的总数也就是最短路的边数+1，因为边权都为1所以最短路的边数也就是最短路的长度。
  实在没思路也没关系，代码加注释如下：
  ps（代码很丑，确定要看吗?）

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`

#include<cstdio>
#include<cstring> 
#define MAXX 10000005
using namespace std;
int n,p,pre[1005],val[1005][1005],k,adda,addb,minl;
bool point[1005];
void Dijkstra_GK_sb() 
{
	pre[1]=0; //1到自己为0.
	for(int i=2;i<=p;i++)
	{
		pre[i]=val[1][i];//1到各点的权值。
	}
	point[1]=true;//从1开始。
	for(int i=1;i<=p-1;i++)
	{
		k=1;//k=1，WA血的教训。
		minl=MAXX;
		for(int ii=1;ii<=p;ii++)
			if(!point[ii]&&pre[ii]<minl)
			{
				minl=pre[ii];
				k=ii;
			}
		    point[k]=true;//表示搜过了。
		for(int j=1;j<=p;j++)
		{
			if(!point[j]&&pre[j]>pre[k]+val[k][j])//如果到一个没搜过的点j，且1到点j的权大于1到k的权加k到j点的权。
				pre[j]=pre[k]+val[k][j];//那么走短的路。
			}
		}
	}
}
int main()
{
	scanf("%d%d",&n,&p);
	memset(point,false,sizeof(point));
	for(int i=1;i<=p;i++)
	    for(int j=1;j<=p;j++)
	    {
	    	val[i][j]=MAXX; 
		}//初始化。
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&adda,&addb);
		val[adda][addb]=1;//边权赋为1.
	}
    Dijkstra_GK_sb(); //dijkstra（请忽略GK_sb）
    if(pre[p]==MAXX)
    {
    	printf("-1\n");
	}
	else
	{
		printf("%d\n",pre[p]+1); //因为边权都为1，所以pre[p]为边数，点数也就为边数+1.
	}
	return 0;
}

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好了，如果代码觉得眼睛可以接受的话，那么谢谢各位神犇的观看。