最短路和次短路的条数(dijstra算法或spfa算法)POJ3463
http://poj.org/problem?id=3463
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 7252 | Accepted: 2581 |
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
题意:给出一个有向图,起点和终点,然后询问最短路和次短路比最短路大1的总条数;
第一种:dijstra
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include"stdio.h"
#include"string.h"
#include"iostream"
#include"map"
#include"string"
#include"queue"
#include"stdlib.h"
#include"algorithm"
#include"vector"
#include"math.h"
#define M 1009
#define eps 1e-5
#define mod 100000000
#define inf 0x3f3f3f3f
using namespace std;
struct node
{
int v,w;
node(int vv,int ww)
{
v=vv;
w=ww;
}
};
vector<node>edge[M];
int dis[M][],vis[M][],num[M][];
void dij(int s,int t,int n)
{
int i,j;
memset(dis,inf,sizeof(dis));
memset(vis,,sizeof(vis));
dis[s][]=;
num[s][]=num[s][]=;
for(i=;i<=n*;i++)
{
int mini=inf;
int u=-;
int flag;
for(j=;j<=n;j++)
{
if(!vis[j][]&&mini>dis[j][])
{
flag=;
mini=dis[j][];
u=j;
}
else if(!vis[j][]&&mini>dis[j][])
{
flag=;
mini=dis[j][];
u=j;
}
}
if(u==-)break;
vis[u][flag]=;
for(j=;j<(int)edge[u].size();j++)
{
int v=edge[u][j].v;
int w=edge[u][j].w;
if(dis[v][]>mini+w)
{
dis[v][]=dis[v][];
dis[v][]=mini+w;
num[v][]=num[v][];
num[v][]=num[u][flag];
}
else if(dis[v][]==mini+w)
{
num[v][]+=num[u][flag];
}
else if(dis[v][]>mini+w)
{
dis[v][]=mini+w;
num[v][]=num[u][flag];
}
else if(dis[v][]==mini+w)
{
num[v][]+=num[u][flag];
}
}
}
int ans;
if(dis[t][]==dis[t][]+)
ans=num[t][]+num[t][];
else
ans=num[t][];
//printf("%d %d %d %d\n",dis[t][0],dis[t][1],num[t][0],num[t][1]);
printf("%d\n",ans);
}
int main()
{
int T,n,m,i;
cin>>T;
while(T--)
{
scanf("%d%d",&n,&m);
for(i=;i<=n;i++)
edge[i].clear();
for(i=;i<m;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
edge[u].push_back(node(v,w));
}
int s,t;
scanf("%d%d",&s,&t);
dij(s,t,n);
}
return ;
}
第二种:spfa
#include"stdio.h"
#include"string.h"
#include"queue"
#include"stdlib.h"
#define M 1009
#define inf 0x3f3f3f3f
using namespace std;
struct Gra
{
int u,v,w,next;
}edge[M*];
int t,head[M],dis[M][],num[M][],use[M][];
void init()
{
t=;
memset(head,-,sizeof(head));
}
void add(int u,int v,int w)
{
edge[t].u=u;
edge[t].v=v;
edge[t].w=w;
edge[t].next=head[u];
head[u]=t++;
}
struct node
{
int v,flag,dis;
node(){}
node(int v,int dis,int flag)
{
this->v=v;
this->dis=dis;
this->flag=flag;
}
bool operator<(const node &a)const
{
return dis>a.dis;
}
};
void dij(int s,int n)
{
priority_queue<node>q;
memset(dis,inf,sizeof(dis));
memset(use,,sizeof(use));
dis[s][]=;
num[s][]=;
q.push(node(s,,));
while(!q.empty())
{
node cur=q.top();
int u=cur.v;
int flag=cur.flag;
q.pop();
if(use[u][flag])continue;
use[u][flag]=;
for(int i=head[u];~i;i=edge[i].next)
{
int v=edge[i].v;
if(dis[v][]>dis[u][flag]+edge[i].w)
{
dis[v][]=dis[v][];
dis[v][]=dis[u][flag]+edge[i].w;
num[v][]=num[v][];
num[v][]=num[u][flag];
q.push(node(v,dis[v][],));
q.push(node(v,dis[v][],));
}
else if(dis[v][]==dis[u][flag]+edge[i].w)
{
num[v][]+=num[u][flag];
}
else if(dis[v][]>dis[u][flag]+edge[i].w)
{
dis[v][]=dis[u][flag]+edge[i].w;
num[v][]=num[u][flag];
q.push(node(v,dis[v][],));
}
else if(dis[v][]==dis[u][flag]+edge[i].w)
{
num[v][]+=num[u][flag];
}
}
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,m;
scanf("%d%d",&n,&m);
init();
for(int i=;i<=m;i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
add(a,b,c);
}
int ss,tt;
scanf("%d%d",&ss,&tt);
dij(ss,n);
int ans=;
ans+=num[tt][];
if(dis[tt][]+==dis[tt][])
ans+=num[tt][];
printf("%d\n",ans);
}
return ;
}
最短路和次短路的条数(dijstra算法或spfa算法)POJ3463的更多相关文章
- Bellman-ford算法、SPFA算法求解最短路模板
Bellman-ford 算法适用于含有负权边的最短路求解,复杂度是O( VE ),其原理是依次对每条边进行松弛操作,重复这个操作E-1次后则一定得到最短路,如果还能继续松弛,则有负环.这是因为最长的 ...
- 图论之最短路算法之SPFA算法
SPFA(Shortest Path Faster Algorithm)算法,是一种求最短路的算法. SPFA的思路及写法和BFS有相同的地方,我就举一道例题(洛谷--P3371 [模板]单源最短路径 ...
- hdu1688(dijkstra求最短路和次短路)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1688 题意:第k短路,这里要求的是第1短路(即最短路),第2短路(即次短路),以及路径条数,最后如果最 ...
- HDU 1688 Sightseeing&HDU 3191 How Many Paths Are There(Dijkstra变形求次短路条数)
Sightseeing Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tota ...
- HDU 3416 Marriage Match IV (求最短路的条数,最大流)
Marriage Match IV 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/Q Description Do not si ...
- HDU 3191 次短路长度和条数
http://www.cnblogs.com/wally/archive/2013/04/16/3024490.html http://blog.csdn.net/me4546/article/det ...
- POJ---3463 Sightseeing 记录最短路和次短路的条数
Sightseeing Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9247 Accepted: 3242 Descr ...
- HDU3191 【输出次短路条数】
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3191 How Many Paths Are There Time Limit: 2000/1000 M ...
- HDU 1688 Sightseeing 【输出最短路+次短路条数】
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1688 题目大意:给n个点,m条有向边.再给出起点s, 终点t.求出s到t的最短路条数+次短路条数. 思 ...
随机推荐
- phpstudy+php5.2+mssql2008
我勒个去.... <?php $server ="XEJMZWMDIXE9CIJ"; //服务器IP地址,如果是本地,可以写成localhost $uid ="&q ...
- tensorflow中slim模块api介绍
tensorflow中slim模块api介绍 翻译 2017年08月29日 20:13:35 http://blog.csdn.net/guvcolie/article/details/77686 ...
- (转)ffplay的音视频同步分析之视频同步到音频
以前工作中参与了一些音视频程序的开发,不过使用的都是芯片公司的SDK,没有研究到更深入一层,比如说音视频同步是怎么回事.只好自己抽点时间出来分析开源代码了,做音视频编解码的人都知道ffmp ...
- 巧用JS中的join方法操作字符串
1.将数组的元素组起一个字符串,以separator为分隔符,省略的话则用默认用逗号为分隔符 /** *把数组转换成特定符号分割的字符串 */ function arrayToString(arr,s ...
- R语言低级绘图函数-text
text函数用来在一张图表上添加文字,只需要指定对应的x和y坐标,以及需要添加的文字内容就可以了 基本用法: plot(1:5, 1:5, xlim = c(0,6), ylim = c (0,6), ...
- HTML input只能输入数字
onkeyup="this.value=this.value.replace(/[^0-9]/g,'')" onafterpaste="this.value=this.v ...
- ubuntu12.04 修改登陆用户 为root
Ubuntu 12.04默认是不允许root登录的,在登录窗口只能看到普通用户和访客登录.以普通身份登陆Ubuntu后我们需要做一些修改,普通用户登录后,修改系统配置文件需要切换到超级用户模式,在终端 ...
- 继承MonoBehaviour类的优缺点和相关报错
Unity3D文档里虽然说所有脚本继承MonoBehaviour类,但如果你想自定义类,就可以不用继承MonoBehaviour,但是这个类只能调用其中的方法和属性,无法拖到场景的物体中使用. 所有从 ...
- 解决导入protobuf源代码Unity报错的问题
将源代码导入Assets目录后, unity引擎会出现以下报错: 解决办法: 在 unity项目Assets目录中创建smcs.rsp文件,内容为-unsafe,其作用为可编译不安全代码. 然 ...
- [extjs] ExtJs4.2 Form 表单提交
基本代码: <script> Ext.onReady(function(){ Ext.create('Ext.form.Panel', { title: '登录', bodyPadding ...