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); }