HDU 4014 Jimmy’s travel plan(图计数)
Jimmy’s travel plan
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 341 Accepted Submission(s): 58
lives in a huge kingdom which contains lots of beautiful cities. He
also loves traveling very much, and even would like to visit each city
in the country. Jaddy, his secretary, is now helping him to plan the
routes, however, Jaddy suddenly find that is quite a tough task because
it is possible for Jimmy to ask route’s information toward any city.
What was worth? Jaddy has to response for queries about the distance
information nearly between any pair of cities due to the undeterminable
starting city which Jimmy is living in when he raises a query. Because
of the large scale of the whole country, Jaddy feel hopeless to archive
such an impossible job, however, in order to gratify his manager, Jaddy
is now looking forward to your assistance.
There might be good news
about Jaddy’s work: since Jimmy is very lazy and would not like to
travel to a destination whose distance between the original city is
larger than TWO. That means only one intermediate city among the route
is acceptable (Apparently, all the connecting paths between any two
cities, if exists, have the same length as ONE). But don’t be fooled:
Jimmy also needs to know that how many alternative different routes are
available so that he can have more options. In particular two routes
were named as different if and only if there is at least one path in the
two routes is distinguishable, moreover, if more than one paths exist
between a particular pair of cities, they are considered as distinct.
has multiple test cases. The first line of the input has a single
integer T indication the number of test cases, then each test case
following. For each test case, the first line contains two integers N
and M indication the number of cities and paths in the country. Then M
lines are following, each line contains a pair of integers A and B,
separated by space, denoting an undirected path between city A and city
B, all the cities are numbered from 1 to N. Then a new line contains a
single integer Q, which means there are Q queries following. Each query
contains a couple of integers A and B which means querying the distance
and number of shortest routes between city A and B, each query occupy a
single line separately.
All the test cases are separated by a single blank line.
You can assume that N, Q <= 100000, M <= 200000.
each test case, firstly output a single line contains the case number,
then Q lines for the response to queries with the same order in the
input. For each query, if there exists at least one routes with length
no longer than TWO, then output two integer separated by a single space,
the former is the distance (shortest) of routes and the later means how
many different shortest routes Jimmy can choose; otherwise, output a
single line contains “The pair of cities are not connected or too far
away.” (quotes for clarifying). See the sample data carefully for
further details.
#include<bits/stdc++.h>
using namespace std ;
const int N = ;
const int M = ;
const long long K = ;
int n , m ;
struct node {
int u , v ;
long long cnt ;
bool operator < ( const node &a ) const {
if( u != a.u ) return u < a.u ;
else return v < a.v ;
}
} e[M] ;
int tot ;
int in[N] ;
const int HASH = ;
struct HASHMAP {
long long key[N] , f[N] ;
int head[HASH] , next[N] , size ;
void init() {
memset( head , - , sizeof head ) ;
size = ;
}
void insert( int u , int v , long long ans ) {
long long KEY = K * u + v ;
int c = KEY % HASH ;
f[size] = ans ;
key[size] = KEY ;
next[size] = head[c] ;
head[c] = size++;
}
long long find( int u , int v ) {
long long KEY = K * u + v ;
int c = KEY % HASH ;
for( int i = head[c] ; ~i ; i = next[i] ) {
if( key[i] == KEY ) return f[i] ;
}
return - ;
}
} mp ; long long find( int u , int v ) {
int l = , r = m - ;
// cout << "l :" << l << " r : " << r << endl ;
if( l > r ) return - ;
if( u > e[r].u || ( u == e[r].u && v > e[r].v ) ) return - ;
if( v < e[l].v || ( v == e[l].v && u < e[l].u ) ) return - ;
while( l <= r ) {
int mid = (l+r)>>;
if( u < e[mid].u ) {
r = mid - ;
}
else if( u == e[mid].u ) {
if( v < e[mid].v ) r = mid - ;
else if( v > e[mid].v ) l = mid + ;
else return e[mid].cnt ;
}
else {
l = mid + ;
}
}
return - ;
}
vector< pair<int,long long> >g[N]; void Work() {
mp.init();
int q ; cin >> q ;
while( q-- ) {
int u , v ; cin >> u >> v ;
u-- , v-- ;
if( u > v ) swap( u , v ) ;
else if( u == v ) { cout << "0 1"<< endl ; continue ; }
long long ans = find( u , v ) ;
if( ans != - ) {
cout << "1 " << ans << endl ;
} else {
ans = mp.find( u , v ) ;
if( ans == - ) {
ans = ;
if( in[u] > in[v] ) swap( u , v ) ;
for( int i = ; i < g[u].size() ; ++i ) {
int mid = g[u][i].first ;
long long cnt1 = g[u][i].second , cnt2 = find( min(mid,v) , max(mid,v) );
if( cnt2 != - ) ans += cnt1 * cnt2 ;
}
if( u > v ) swap( u , v ) ;
mp.insert( u , v , ans ) ;
}
if( ans ) cout << "2 " << ans << endl ;
else cout << "The pair of cities are not connected or too far away." << endl ;
}
}
} void Gao() {
memset( in , , sizeof in );
for( int i = ; i < n ; ++i ) g[i].clear();
cin >> n >> m ;
if( !m ) return ;
for( int i = ; i < m ; ++i ) {
int x , y ; cin >> x >> y ;
x-- , y-- ;
if( x > y ) swap( x , y ) ;
e[i].u = x , e[i].v = y , e[i].cnt = ;
in[x]++ , in[y]++ ;
}
sort( e , e + m ) ;
tot = ;
for( int i = ; i < m ; ++i ) {
if( e[tot-].u == e[i].u && e[tot-].v == e[i].v ) {
e[tot-].cnt++ ;
} else {
e[tot++] = e[i] ;
}
}
m = tot ;
for( int i = ; i < m ; ++i ) {
g[e[i].u].push_back( make_pair( e[i].v,e[i].cnt) ) ;
g[e[i].v].push_back( make_pair( e[i].u,e[i].cnt) ) ;
}
} int Run() {
int _ , cas = ; cin >> _ ;
while( _-- ) {
cout << "Case #" << cas++ << ":" << endl ;
Gao(); Work();
}
return ;
}
int main() {
ios::sync_with_stdio();
return Run();
}
HDU 4014 Jimmy’s travel plan(图计数)的更多相关文章
- hdu 4885 TIANKENG’s travel(bfs)
题目链接:hdu 4885 TIANKENG's travel 题目大意:给定N,L,表示有N个加油站,每次加满油能够移动距离L,必须走直线,可是能够为斜线.然后给出sx,sy,ex,ey,以及N个加 ...
- PAT-1030 Travel Plan (30 分) 最短路最小边权 堆优化dijkstra+DFS
PAT 1030 最短路最小边权 堆优化dijkstra+DFS 1030 Travel Plan (30 分) A traveler's map gives the distances betwee ...
- PAT1030 Travel Plan (30)---DFS
(一)题意 题目链接:https://www.patest.cn/contests/pat-a-practise/1030 1030. Travel Plan (30) A traveler's ma ...
- 有标号的DAG图计数1~4
前言 我什么都不会,菜的被关了起来. 有标号的DAG图I Solution 考虑递推,设\(f_i\)表示i个点的答案,显然这个东西是可以组合数+容斥递推? 设\(f_i\)表示i个点的答案,我们考虑 ...
- PAT 1030 Travel Plan[图论][难]
1030 Travel Plan (30)(30 分) A traveler's map gives the distances between cities along the highways, ...
- 1030 Travel Plan (30 分)
1030 Travel Plan (30 分) A traveler's map gives the distances between cities along the highways, toge ...
- [图算法] 1030. Travel Plan (30)
1030. Travel Plan (30) A traveler's map gives the distances between cities along the highways, toget ...
- PAT_A1030#Travel Plan
Source: PAT A1030 Travel Plan (30 分) Description: A traveler's map gives the distances between citie ...
- PAT 甲级 1030 Travel Plan (30 分)(dijstra,较简单,但要注意是从0到n-1)
1030 Travel Plan (30 分) A traveler's map gives the distances between cities along the highways, to ...
随机推荐
- re模块下的的常用方法
引入模块: import re 1.查找findall 匹配所有,每一项都是列表中的一个元素 ret=re.findall("\d+","sjkhk172按实际花费9 ...
- 【leetcode】1137. N-th Tribonacci Number
题目如下: The Tribonacci sequence Tn is defined as follows: T0 = 0, T1 = 1, T2 = 1, and Tn+3 = Tn + Tn+1 ...
- 接口返回[object,Object]解决方法
1.我们请求接口时有时会返回[object,Object],[object,Object],[object,Object].... 这个我们使用JSON.stringfity(data),就可以解决.
- Java——容器(Auto-boxing/unboxing)
[打包/解包] 在Map中需要增加一个数值时,需要new一个对象出来,输出后又得进行强制类型转换,这就造成不便.在JDK1.5中使用Map接口提供了一种新的机制. 在合适的时机自动打包/解包(在J ...
- 详细讲解Android中的AbsListView的源码
不知道各位童鞋们在开发的过程中有没有感兴趣过ListView是如何实现的呢?其实本身ListView的父类AbsListView才是关键,但是如果大家看过源码的话,会发现AbsListView将近70 ...
- echarts曲线图
drawLeftLine(){ let drawLine = echarts.init(document.getElementById('data-left-middle-table-wrap')); ...
- 2017ICPC南宁补题
https://www.cnblogs.com/2462478392Lee/p/11650548.html https://www.cnblogs.com/2462478392Lee/p/116501 ...
- 小程序中css3实现优惠券
效果如下: css3实现优惠券 知识储备 颜色渐变 linear-gradient() css伪类 :before :after index.wxss .app { /* padding: 20rpx ...
- 【MySQL】 知识点记录
0. 定位和排查问题的常用语句 查询 正在执行的事务(这个输出有事物状态表明是否等待锁):SELECT * FROM information_schema.INNODB_TRX 查看正在锁的事务:SE ...
- bootstrap插件bootstrap-select使用demo
插件bootstrap-select官网 : https://developer.snapappointments.com/bootstrap-select/ bootstrap-select插件: ...