[USACO08OCT]牧场散步Pasture Walking

Posted 2020-10-06 心之所向 素履以往

                                  [USACO08OCT]牧场散步Pasture Walking

 

题目描述

The N cows (2 <= N <= 1,000) conveniently numbered 1..N are grazing among the N pastures also conveniently numbered 1..N. Most conveniently of all, cow i is grazing in pasture i.

Some pairs of pastures are connected by one of N-1 bidirectional walkways that the cows can traverse. Walkway i connects pastures A_i and B_i (1 <= A_i <= N; 1 <= B_i <= N) and has a length of L_i (1 <= L_i <= 10,000).

The walkways are set up in such a way that between any two distinct pastures, there is exactly one path of walkways that travels between them. Thus, the walkways form a tree.

The cows are very social and wish to visit each other often. Ever in a hurry, they want you to help them schedule their visits by computing the lengths of the paths between 1 <= L_i <= 10,000 pairs of pastures (each pair given as a query p1,p2 (1 <= p1 <= N; 1 <= p2 <= N).

POINTS: 200

有N(2<=N<=1000)头奶牛，编号为1到W，它们正在同样编号为1到N的牧场上行走.为了方 便，我们假设编号为i的牛恰好在第i号牧场上.

有一些牧场间每两个牧场用一条双向道路相连，道路总共有N - 1条，奶牛可以在这些道路 上行走.第i条道路把第Ai个牧场和第Bi个牧场连了起来(1 <= A_i <= N; 1 <= B_i <= N)，而它的长度 是 1 <= L_i <= 10,000.在任意两个牧场间，有且仅有一条由若干道路组成的路径相连.也就是说，所有的道路构成了一棵树.

奶牛们十分希望经常互相见面.它们十分着急，所以希望你帮助它们计划它们的行程，你只 需要计算出Q(1 < Q < 1000)对点之间的路径长度•每对点以一个询问p1,p2 (1 <= p1 <= N; 1 <= p2 <= N). 的形式给出.

输入输出格式

输入格式：

 

• Line 1: Two space-separated integers: N and Q

• Lines 2..N: Line i+1 contains three space-separated integers: A_i, B_i, and L_i

• Lines N+1..N+Q: Each line contains two space-separated integers representing two distinct pastures between which the cows wish to travel: p1 and p2

 

输出格式：

 

• Lines 1..Q: Line i contains the length of the path between the two pastures in query i.

 

输入输出样例

输入样例#1：

4 2 
2 1 2 
4 3 2 
1 4 3 
1 2 
3 2 

输出样例#1：

2 
7 

说明

Query 1: The walkway between pastures 1 and 2 has length 2.

Query 2: Travel through the walkway between pastures 3 and 4, then the one between 4 and 1, and finally the one between 1 and 2, for a total length of 7.

 

考察对floyd的理解

10000条边 10000次询问 

显然我们要尽量做到O(1) 查询

n<=250 显然 考虑 Floyd

n^3 是可以承受的

最短路明显好求 

断点k 并不一定是当前的最大点权 

这样比较的k就不是路径上除i，j之外的点权最大值

所以要枚举最大点权  

但是n^4 代价太大 所以先对 点权排序 

为了不让 点权的标号混乱 统一用新的点替换 旧点 

做到O(1)枚举 最大点权 

k既是断点 也是当前的最大点权 

跑一遍floyd就好了

 

 1 #include <queue>
 2 #include <cstdio>
 3 #include <cctype>
 4 #include <cstring>
 5 #include <algorithm>
 6 
 7 const int MAXN=500;
 8 const int INF=0x3f3f3f3f;
 9 
10 int n,m,K;
11 
12 int a[MAXN],f[MAXN][MAXN],map[MAXN][MAXN];
13 
14 struct node {
15     int val;
16     int id;
17     node() {}
18     node(int val,int id):val(val),id(id) {}
19     friend bool operator < (node a,node b) {
20         return a.val<b.val;
21     }
22 };
23 node e[MAXN];
24 
25 inline void read(int&x) {
26     int f=1;register char c=getchar();
27     for(x=0;!isdigit(c);c==‘-‘&&(f=-1),c=getchar());
28     for(;isdigit(c);x=x*10+c-48,c=getchar());
29     x=x*f;
30 }
31 
32 inline int min(int a,int b) {return a<b?a:b;}
33 
34 inline int max(int a,int b) {return a<b?b:a;}
35 
36 int hh() {
37 //    freopen("fee.in","r",stdin);
38 //    freopen("fee.out","w",stdout);
39     read(n);read(m);read(K);
40     memset(f,INF,sizeof f);
41     memset(map,INF,sizeof map);
42     for(int i=1;i<=n;++i) read(e[i].val),e[i].id=i;
43     std::sort(e+1,e+1+n);
44     for(int i=1;i<=n;++i) a[e[i].id]=i;
45     for(int x,y,z,i=1;i<=m;++i) {
46         read(x);read(y);read(z);
47         x=a[x];y=a[y];
48         f[x][y]=min(f[x][y],z);
49         f[y][x]=f[x][y];
50     }
51 //    for(int i=1;i<=n;++i) f[i][i]=0;
52     for(int k=1;k<=n;++k)
53       for(int i=1;i<=n;++i)
54         for(int j=1;j<=n;++j) {
55             f[i][j]=min(f[i][j],f[i][k]+f[k][j]);
56             map[i][j]=min(map[i][j],f[i][j]+max(e[k].val,max(e[i].val,e[j].val)));
57         }
58     for(int S,T,i=1;i<=K;++i) {
59         read(S);read(T);
60         if(map[a[S]][a[T]]==INF) printf("-1\n");
61         else printf("%d\n",map[a[S]][a[T]]);
62     }
63     return 0;
64 }
65 
66 int sb=hh();
67 int main(int argc,char**argv) {;}

代码