「BZOJ1674」[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
OUTPUT DETAILS:
The cows possess 4 objects in total: first they trade object 1 for
object 3, then object 3 for object 2, then object 2 for object 5.
OUTPUT DETAILS:
The cows possess 4 objects in total: first they trade object 1 for
object 3, then object 3 for object 2, then object 2 for object 5.
题解
直接最短路T T
去搞了个dijkstra+heap的模板玩玩
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#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<queue> #include<vector> #define ll long long #define inf 1000000000 #define pa pair<int,int> using namespace std; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,m,cnt; int dis[1005],last[1005]; bool vis[1005]; struct edge{int to,next,v;}e[50005]; void insert(int u,int v,int w) { e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;e[cnt].v=w; } void dijkstra() { priority_queue<pa,vector<pa>,greater<pa> > q; for(int i=1;i<=n;i++)dis[i]=inf; dis[1]=0; q.push(make_pair(0,1)); while(!q.empty()) { int now=q.top().second;q.pop(); if(vis[now])continue;vis[now]=1; for(int i=last[now];i;i=e[i].next) { int to=e[i].to; if(dis[to]>dis[now]+e[i].v) { dis[to]=dis[now]+e[i].v; q.push(make_pair(dis[to],to)); } } } } int main() { m=read();n=read(); for(int i=1;i<=m;i++) { int u=read(),v=read(),w=1; insert(u,v,w); } dijkstra(); if(dis[n]==inf)puts("-1"); else printf("%d",dis[n]+1); return 0; } |
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