1674: [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.
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号货物.
1 3 //1号星球,希望得到1号货物,将给你3号货物
3 2
2 3
3 1
2 5
5 4
1 3 //1号星球,希望得到1号货物,将给你3号货物
3 2
2 3
3 1
2 5
5 4
Sample Output
4
思路：
最短路裸题， 把输入看成点和边， 边权均为1 跑一遍Dij即可， 我是反着跑的， 答案一样， 注意反向加边 另外用的配对堆以期实现nlogn的复杂度
#include <cstdio>
#include <ext/pb_ds/priority_queue.hpp>
#include <iostream>
#include <cstring>
#define ll long long
/*#define __attribute__((optimize("-O2")))*/
#define mp(a, b) make_pair(a, b)
using namespace std;
using namespace __gnu_pbds;
#define heap __gnu_pbds::priority_queue<pair<ll, int> >
const ll INF = 1e15;
const int N = 1000001, M = 1000001;
struct node {
int to, val, next;
}e[M];
int n, m, head[N], cnt;
ll f[N];
heap q;
inline char nc() {
static char buf[100000], *p1, *p2;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin), p1==p2)?EOF:*p1++;
}
inline int read() {
char c=nc();int x=0;
while(!isdigit(c))c=nc();
while(isdigit(c)){x=(x<<3)+(x<<1)+(c^'0');c=nc();}
return x;
}
inline void add(int x, int y, int z) {
e[++cnt].to = y;
e[cnt].next = head[x];
head[x] = cnt;
e[cnt].val = z;
}
heap::point_iterator id[N];
int main() {
/*#ifndef ONLINE_JUDGE
freopen("3040.in", "r", stdin);
#endif*/
int n=read(), m=read(); //m -= read(); read();read();read();read();read();
register int i;
int x, y;
for(i=1;i<=n;i++) {
x=read(), y=read();
add(y, x, 1);
}
for(i=1;i<m;i++) f[i] = INF;
id[m] = q.push(mp(0, m));
while(!q.empty()) {
int u = q.top().second;q.pop();
for(i=head[u];i;i=e[i].next) {
if(f[e[i].to]>f[u]+e[i].val) {
f[e[i].to] = f[u] + e[i].val;
if(id[e[i].to]!=0) q.modify(id[e[i].to], mp(-f[e[i].to], e[i].to));
else id[e[i].to] = q.push(mp(-f[e[i].to], e[i].to));
}
}
}
printf("%lld", (f[1]+1)>1e9?-1:f[1]+1);
}