Bzoj 1674: [Usaco2005]Part Acquisition Dij / Spfa

1674: [Usaco2005]Part Acquisition


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.


* 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.


* 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

Sample Output



 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;
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;
    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("", "r", stdin);
    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.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);


posted @ 2018-06-30 16:08  TOBICHI_ORIGAMI  阅读(62)  评论(0编辑  收藏  举报