[Usaco2005]Part Acquisition

Description

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.

有n个星球，开始你的手中有1号物品，每个星球会帮你把物品a换成物品b，问你要得到物品m最少要换几次，如果怎么样都不能换到，输出-1

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 //6个星球,希望得到5,开始时你手中有1号物品.

1 3 //1号星球，帮你把1号物品换成3号

3 2

2 3

3 1

2 5

5 4

Sample Output

4

这题裸的最短路，什么都不用想

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline int read(){
    int x=0,f=1;char ch=getchar();
    for (;ch<‘0‘||ch>‘9‘;ch=getchar())  if (ch==‘-‘)    f=-1;
    for (;ch>=‘0‘&&ch<=‘9‘;ch=getchar())    x=(x<<1)+(x<<3)+ch-‘0‘;
    return x*f;
}
inline void print(int x){
    if (x>=10)     print(x/10);
    putchar(x%10+‘0‘);
}
const int N=5e4;
int pre[N+10],now[N+10],child[N+10];
int h[N+10],dis[N+10];
bool vis[N+10];
int tot,n,End,root=1;
void join(int x,int y){pre[++tot]=now[x],now[x]=tot,child[tot]=y;}
void SPFA(int x){
    int head=0,tail=1;
    memset(dis,63,sizeof(dis));
    h[1]=x,vis[x]=1,dis[x]=1;
    while (head!=tail){
        if (++head>N)   head=1;
        int Now=h[head];
        for (int p=now[Now],son=child[p];p;p=pre[p],son=child[p])
            if (dis[son]>dis[Now]+1){
                dis[son]=dis[Now]+1;
                if (!vis[son]){
                    if (++tail>N)   tail=1;
                    h[tail]=son,vis[son]=1;
                }
            }
        vis[Now]=0;
    }
}
int main(){
    n=read(),End=read();
    for (int i=1,x,y;i<=n;i++)  x=read(),y=read(),join(x,y);
    SPFA(root);
    dis[End]!=inf?printf("%d\n",dis[End]):printf("-1\n");
    return 0;
}

原文地址：https://www.cnblogs.com/Wolfycz/p/8411137.html