POJ3613 Cow Relays

时间：2016-07-12 22:55:58 阅读：232 评论：0 收藏：0 [点我收藏+]

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Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6726		Accepted: 2626

Description

For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout the pasture.

Each trail connects two different intersections (1 ≤ I_1i ≤ 1,000; 1 ≤ I_2i ≤ 1,000), each of which is the termination for at least two trails. The cows know the length_i of each trail (1 ≤ length_i ≤ 1,000), the two intersections the trail connects, and they know that no two intersections are directly connected by two different trails. The trails form a structure known mathematically as a graph.

To run the relay, the N cows position themselves at various intersections (some intersections might have more than one cow). They must position themselves properly so that they can hand off the baton cow-by-cow and end up at the proper finishing place.

Write a program to help position the cows. Find the shortest path that connects the starting intersection (S) and the ending intersection (E) and traverses exactly N cow trails.

Input

* Line 1: Four space-separated integers: N, T, S, and E
* Lines 2..T+1: Line i+1 describes trail i with three space-separated integers: length_i , I_1i , and I_2i

Output

* Line 1: A single integer that is the shortest distance from intersection S to intersection E that traverses exactly N cow trails.

Sample Input

Sample Output

Source

USACO 2007 November Gold

经过n条边的最短路径。

好像可以贪心，先找到一条最短路，然后随便找个最小环绕圈圈，我没有尝试。

使用了标准解法，矩阵乘法+倍增floyd

floyd是枚举k点，更新i到j的最短路径，更新后实际上就是从原本的走一条路到达更新成了走两条路到达。

根据这个思想，我们可以倍增跑floyd，就能得出走2^k条边从i到j的最短路径。

代码如下：（a[i]里存的就是走2^i条边的最短路邻接矩阵)

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 using namespace std;
 7 const int mxn=1000;
 8 bool vis[mxn];
 9 int n;
10 int fa[mxn];
11 int in[mxn],out[mxn];
12 void init(int n){
13     for(int i=1;i<=n;i++)fa[i]=i;
14 }
15 int find(int x){
16     if(fa[x]==x)return x;
17     else return fa[x]=find(fa[x]);
18 }
19 char s[1100];
20 int main(){
21     int T;
22     scanf("%d",&T);
23     while(T--){
24         memset(vis,0,sizeof vis);
25         memset(in,0,sizeof in);
26         memset(out,0,sizeof out);
27         scanf("%d",&n);
28         init(28);
29         int i,j;
30         for(i=1;i<=n;i++){
31             scanf("%s",s);
32             int u=s[0]-‘a‘+1;
33             int v=s[strlen(s)-1]-‘a‘+1;
34             in[v]++;out[u]++;
35             vis[u]=vis[v]=1;
36             int x=find(u),y=find(v);
37             if(x!=y)fa[y]=x;
38         }
39         int st=0,ed=0;bool flag=0;
40         int cnt=0;
41         for(i=1;i<=26;i++){
42             if(vis[i] && find(i)==i){
43                 cnt++;
44             }
45             if(out[i]!=in[i]){//出度不等于入度时判定 
46                 if(out[i]==in[i]+1){
47                     if(!st)st=i;
48                     else flag=1;
49                 }
50                 else if(in[i]==out[i]+1){
51                     if(!ed)ed=i;
52                     else flag=1;
53                 }
54                 else flag=1;
55             }
56         }
57         if(cnt!=1){//非连通特判 
58             printf("The door cannot be opened.\n");
59             continue;
60             }
61         if(flag==1)printf("The door cannot be opened.\n");
62         else printf("Ordering is possible.\n");
63     }
64     return 0;
65 }

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6726		Accepted: 2626