POJ 3463 Sightseeing (次短路经数)
Sightseeing
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions:10005 | Accepted: 3523 |
Description
Tour operator Your Personal Holiday organises guided bus trips across the Benelux. Every day the bus moves from one city S to another city F. On this way, the tourists in the bus can see the sights alongside the route travelled. Moreover, the bus makes a number of stops (zero or more) at some beautiful cities, where the tourists get out to see the local sights.
Different groups of tourists may have different preferences for the sights they want to see, and thus for the route to be taken from S to F. Therefore, Your Personal Holiday wants to offer its clients a choice from many different routes. As hotels have been booked in advance, the starting city S and the final city F, though, are fixed. Two routes from S to F are considered different if there is at least one road from a city A to a city B which is part of one route, but not of the other route.
There is a restriction on the routes that the tourists may choose from. To leave enough time for the sightseeing at the stops (and to avoid using too much fuel), the bus has to take a short route from S to F. It has to be either a route with minimal distance, or a route which is one distance unit longer than the minimal distance. Indeed, by allowing routes that are one distance unit longer, the tourists may have more choice than by restricting them to exactly the minimal routes. This enhances the impression of a personal holiday.

For example, for the above road map, there are two minimal routes from S = 1 to F = 5: 1 → 2 → 5 and 1 → 3 → 5, both of length 6. There is one route that is one distance unit longer: 1 → 3 → 4 → 5, of length 7.
Now, given a (partial) road map of the Benelux and two cities S and F, tour operator Your Personal Holiday likes to know how many different routes it can offer to its clients, under the above restriction on the route length.
Input
The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:
One line with two integers N and M, separated by a single space, with 2 ≤ N ≤ 1,000 and 1 ≤ M ≤ 10, 000: the number of cities and the number of roads in the road map.
M lines, each with three integers A, B and L, separated by single spaces, with 1 ≤ A, B ≤ N, A ≠ B and 1 ≤ L ≤ 1,000, describing a road from city A to city B with length L.
The roads are unidirectional. Hence, if there is a road from A to B, then there is not necessarily also a road from B to A. There may be different roads from a city A to a city B.
One line with two integers S and F, separated by a single space, with 1 ≤ S, F ≤ N and S ≠ F: the starting city and the final city of the route.
There will be at least one route from S to F.
Output
For every test case in the input file, the output should contain a single number, on a single line: the number of routes of minimal length or one distance unit longer. Test cases are such, that this number is at most 109 = 1,000,000,000.
Sample Input
2
5 8
1 2 3
1 3 2
1 4 5
2 3 1
2 5 3
3 4 2
3 5 4
4 5 3
1 5
5 6
2 3 1
3 2 1
3 1 10
4 5 2
5 2 7
5 2 7
4 1
Sample Output
3
2
Hint
The first test case above corresponds to the picture in the problem description.
思路
一开始我想了一下,用A*好像可以做,但是超了内存,看了一下discuss,原因应该是不断入队造成的,这让我感到十分无奈,毕竟刚刚才用a*过了一题。
具体的东西我写在注释里了,注意次短路的入队操作!
#include<iostream>
#include<vector>
#include<queue>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const ll inf = 210000000000000;
int n,m,s,t;
vector<int>u[1024];
vector<ll>w[1024];
ll dis[1024][2];
//虽然按照我的推理,这个题不会爆int ,可是不用int,就是会wa;
bool book[1024][2];
ll ans[1024][2];
struct node
{
int num;
ll dis;
int flag;
bool operator<(const node x)const
{
return x.dis<dis;
}
}; ll Dijkstra()
{
priority_queue<node>q;
node exa;
dis[s][0]=0;ans[s][0]=1;
q.push(node{s,0,0});
while(!q.empty()){
exa=q.top();q.pop();
int st = exa.num;//当前起点
ll diss=exa.dis;//diss表示,当前节点的最短(或次短)
int f=exa.flag;//表示这是最短还是次短
if(book[st][f]){continue;}
book[st][f]=true;
int siz=u[st].size();
for(int i=0;i<siz;i++){
int y=u[st][i];
ll ww=w[st][i];
/*y是下一个节点,ww是路径*/
if(dis[y][0]>diss+ww){//最短的距离可以更新
q.push(node{y,dis[y][0],1});
dis[y][1]=dis[y][0];//下一节点的次短路,就是更新前的最短路
ans[y][1]=ans[y][0];//下一节点的次短路条数,就是更新前的最短路条数
dis[y][0]=diss+ww;//更新最短路
ans[y][0]=ans[st][0];//下一最短路的条数,就是当前节点最短/次短的条数
//这个位置不可能是次短路更新
q.push(node{y,diss+ww,0});
}
else if(dis[y][0]==diss+ww){//下一节点的最短距离,就是当前距离
ans[y][0]+=ans[st][0];
}
else if(dis[y][1]>diss+ww){//次短路可更新
dis[y][1]=diss+ww;//更新次短路
ans[y][1]=ans[st][f];//下一节点的次短路条数,就是更新来源的路径数
q.push(node{y,diss+ww,1});
}
else if(dis[y][1]==diss+ww){
ans[y][1]+=ans[st][f];//下一节点的路径数,加上当前节点的路径数
}
}
}
if(dis[t][1]==dis[t][0]+1){
return ans[t][0]+ans[t][1];
}
return ans[t][0];
} void init()
{
for(int i=0;i<=n+1;i++){
for(int j=0;j<=1;j++){
dis[i][j]=inf;
}
}
memset(book,0,sizeof(book));
memset(ans,0,sizeof(ans));
for(int i=1;i<=n;i++){
u[i].clear();
w[i].clear();
}
} int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
init();
int x,y;ll z;
for(int i=1;i<=m;i++){
scanf("%d%d%lld",&x,&y,&z);
u[x].push_back(y);
w[x].push_back(z);
}
scanf("%d%d",&s,&t);
printf("%lld\n",Dijkstra());
}
}
妈呀,wa死我了,a个题真不容易
POJ 3463 Sightseeing (次短路经数)的更多相关文章
- poj 3463 Sightseeing(次短路+条数统计)
/* 对dij的再一次理解 每个点依旧永久标记 只不过这里多搞一维 0 1 表示最短路还是次短路 然后更新次数相当于原来的两倍 更新的时候搞一下就好了 */ #include<iostream& ...
- poj 3463 Sightseeing( 最短路与次短路)
http://poj.org/problem?id=3463 Sightseeing Time Limit: 2000MS Memory Limit: 65536K Total Submissio ...
- POJ - 3463 Sightseeing 最短路计数+次短路计数
F - Sightseeing 传送门: POJ - 3463 分析 一句话题意:给你一个有向图,可能有重边,让你求从s到t最短路的条数,如果次短路的长度比最短路的长度多1,那么在加上次短路的条数. ...
- POJ 3463 Sightseeing (次短路)
题意:求两点之间最短路的数目加上比最短路长度大1的路径数目 分析:可以转化为求最短路和次短路的问题,如果次短路比最短路大1,那么结果就是最短路数目加上次短路数目,否则就不加. 求解次短路的过程也是基于 ...
- poj 3463 Sightseeing——次短路计数
题目:http://poj.org/problem?id=3463 当然要给一个点记最短路和次短路的长度和方案. 但往优先队列里放的结构体和vis竟然也要区分0/1,就像把一个点拆成两个点了一样. 不 ...
- POJ 3463 Sightseeing
最短路+次短路(Dijkstra+priority_queue) 题意是要求你找出最短路的条数+与最短路仅仅差1的次短路的条数. 開始仅仅会算最短路的条数,和次短路的长度.真是给次短路条数跪了.ORZ ...
- POJ 3463 Sightseeing 【最短路与次短路】
题目 Tour operator Your Personal Holiday organises guided bus trips across the Benelux. Every day the ...
- POJ 3463 Sightseeing 题解
题目 Tour operator Your Personal Holiday organises guided bus trips across the Benelux. Every day the ...
- POJ 3463 最(次)短路条数
Sightseeing Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9497 Accepted: 3340 Descr ...
随机推荐
- JSP从入门到精通
1. jsp开发环境配置 在windows下配置jsp的开发环境: 假设已经安装好了jdk,下面来配置tomcat 去http://tomcat.apache.org 下载tomcat windows ...
- css元素选择器
css的元素选择器就是html的标签名:
- freemarker -include
在ftl中使用<#include >时 ,页面被强制显示 需要在struts.xml增加跳转type ,或则可以加同一<result-types></result-typ ...
- How to gitignore
git rm -r --cached . git add . git commit -m "remove gitignore cache" git push
- Luogu1137 旅行计划(拓扑排序)
题目传送门 拓扑排序板子题,模拟即可. 代码 #include<cstdio> #include<iostream> #include<cmath> #includ ...
- luogu P1353 【[USACO08JAN]跑步Running】
USACO!!! 唉!无一例外又是母牛(终于知道美国的牧场养什么了) 今天的主人公是一个叫贝茜的公主病母牛(好洋气) 可是她叫什么和我们理解题好像没有什么关系 通过读题我们可以发现她有三个傲娇的地方 ...
- ExaWizards 2019
AB:div 3 AB??? C:div 1 C???场内自闭的直接去看D.事实上是个傻逼题,注意到物品相对顺序不变,二分边界即可. #include<iostream> #include ...
- 二:C#对象、集合、DataTable与Json内容互转示例;
导航目录: Newtonsoft.Json 概述 一:Newtonsoft.Json 支持序列化与反序列化的.net 对象类型: 二:C#对象.集合.DataTable与Json内容互转示例: ...
- A/B HDU - 1576 (exgcd)
要求(A/B)%9973,但由于A很大,我们只给出n(n=A%9973)(我们给定的A必能被B整除,且gcd(B,9973) = 1). Input数据的第一行是一个T,表示有T组数据. 每组数据有两 ...
- Day24-ModelForm操作及验证
Day23内容回顾--缺失,遗憾成狗. 一:Model(2个功能) -数据库操作: -验证,只有一个clean方法可以作为钩子. 二:Form(专门来做验证的)--------根据form里面写的类, ...