Write a program to find the weighted shortest distances from any vertex to a given source vertex in a digraph. It is guaranteed that all the weights are positive.

Format of functions:

void ShortestDist( MGraph Graph, int dist[], Vertex S );

where MGraph is defined as the following:

typedef struct GNode *PtrToGNode;
struct GNode{
int Nv;
int Ne;
WeightType G[MaxVertexNum][MaxVertexNum];
};
typedef PtrToGNode MGraph;

The shortest distance from V to the source S is supposed to be stored in dist[V]. If V cannot be reached from S, store -1 instead.

Sample program of judge:

#include <stdio.h>
#include <stdlib.h> typedef enum {false, true} bool;
#define INFINITY 1000000
#define MaxVertexNum 10 /* maximum number of vertices */
typedef int Vertex; /* vertices are numbered from 0 to MaxVertexNum-1 */
typedef int WeightType; typedef struct GNode *PtrToGNode;
struct GNode{
int Nv;
int Ne;
WeightType G[MaxVertexNum][MaxVertexNum];
};
typedef PtrToGNode MGraph; MGraph ReadG(); /* details omitted */ void ShortestDist( MGraph Graph, int dist[], Vertex S ); int main()
{
int dist[MaxVertexNum];
Vertex S, V;
MGraph G = ReadG(); scanf("%d", &S);
ShortestDist( G, dist, S ); for ( V=0; V<G->Nv; V++ )
printf("%d ", dist[V]); return 0;
} /* Your function will be put here */

Sample Input (for the graph shown in the figure):

7 9
0 1 1
0 5 1
0 6 1
5 3 1
2 1 2
2 6 3
6 4 4
4 5 5
6 5 12
2

Sample Output:

-1 2 0 13 7 12 3
不能抵达的为INFINITY,用过dijkstra算法,最后记得把INFINITY变成-1,dist[S]变成0
代码:
void ShortestDist( MGraph Graph, int dist[], Vertex S )
{
for(int i = ;i < Graph -> Nv;i ++)
dist[i] = Graph -> G[S][i];
int vis[MaxVertexNum] = {};
vis[S] = ;
for(int i = ;i < Graph -> Nv;i ++)
{
int min = INFINITY;
int t = INFINITY;
for(int j = ;j < Graph -> Nv;j ++)
if(!vis[j]&&dist[j] < min)min = dist[j],t = j;
if(min == INFINITY)continue;
vis[t] = ;
for(int j = ;j < Graph -> Nv;j ++)
{
if(!vis[j])
{
if(dist[j] > Graph -> G[t][j] + min)dist[j] = Graph -> G[t][j] + min;
}
}
}
for(int i = ;i < Graph -> Nv;i ++)
if(i == S)dist[i] = ;
else if(dist[i] == INFINITY)dist [i] = -;
}

6-17 Shortest Path [2](25 分)的更多相关文章

  1. 1126 Eulerian Path (25 分)

    1126 Eulerian Path (25 分) In graph theory, an Eulerian path is a path in a graph which visits every ...

  2. 【刷题-PAT】A1126 Eulerian Path (25 分)

    1126 Eulerian Path (25 分) In graph theory, an Eulerian path is a path in a graph which visits every ...

  3. PAT A1126 Eulerian Path (25 分)——连通图,入度

    In graph theory, an Eulerian path is a path in a graph which visits every edge exactly once. Similar ...

  4. 1003 Emergency (25 分)

    1003 Emergency (25 分) As an emergency rescue team leader of a city, you are given a special map of y ...

  5. HDU 4725 The Shortest Path in Nya Graph (最短路 )

    This is a very easy problem, your task is just calculate el camino mas corto en un grafico, and just ...

  6. 1003 Emergency (25分) 求最短路径的数量

    1003 Emergency (25分)   As an emergency rescue team leader of a city, you are given a special map of ...

  7. 1122 Hamiltonian Cycle (25 分)

    1122 Hamiltonian Cycle (25 分) The "Hamilton cycle problem" is to find a simple cycle that ...

  8. HDU - 4725 The Shortest Path in Nya Graph 【拆点 + dijkstra】

    This is a very easy problem, your task is just calculate el camino mas corto en un grafico, and just ...

  9. Shortest Path(hdu5636)

    Shortest Path  Accepts: 40  Submissions: 610  Time Limit: 4000/2000 MS (Java/Others)  Memory Limit: ...

  10. PTA 旅游规划(25 分)

    7-10 旅游规划(25 分) 有了一张自驾旅游路线图,你会知道城市间的高速公路长度.以及该公路要收取的过路费.现在需要你写一个程序,帮助前来咨询的游客找一条出发地和目的地之间的最短路径.如果有若干条 ...

随机推荐

  1. com.sun.image.codec.jpeg在Eclipse中报错的解决办法

    在Eclipse中处理图片,需要引入两个包:import com.sun.image.codec.jpeg.JPEGCodec;import com.sun.image.codec.jpeg.JPEG ...

  2. #C++初学记录 (第一次测试)(A - 复习时间 )

    练习题目一 (A-复习时间) 为了能过个好年,xhd开始复习了,于是每天晚上背着书往教室跑.xhd复习有个习惯,在复习完一门课后,他总是挑一门更简单的课进行复习,而他复习这门课的效率为两门课的难度差的 ...

  3. python面向对象编程基础

    演示了 Python 类与对象的编程基础, 包括属性.方法.继承.组合.动态创建类. python 版本: 2.7.5 class SimpleClass(object): ''' a simple ...

  4. Python3.x:os.path模块

    Python3.x:os.path模块 #返回绝对路径 os.path.abspath(path) #返回文件名 os.path.basename(path) #返回list(多个路径)中,所有pat ...

  5. GreenOpenPaint的实现(四)放大缩小处理滚动事件

    放大缩小看似简单,实际上还是比较复杂的.所以专门拿出来说明. 缩放这块,主要就是处理m_pDoc->m_scalefactor void CGreenOpenPaintView::OnButto ...

  6. Linux下查找大文件,大目录的方法

    查找大文件 //列举出当前目录所有大于800M的文件 find . -type f -size +800M 1 2 第一个方法只用到了一个命令find,它能够帮我们做一些文件查找的操作.它常用的参数有 ...

  7. Java多线程 线程状态及转换 wait sleep yield join

    线程的状态转化关系(1). 新建状态(New):新创建了一个线程对象.(2). 就绪状态(Runnable):线程对象创建后,其他线程调用了该对象的start()方法.该状态的线程位于可运行线程池中, ...

  8. github上fork别人的代码之后,如何保持和原作者同步的更新

    1.从自己fork之后的版本库clone $  git clone -o chucklu https://github.com/chucklu/Hearthstone-Deck-Tracker.git ...

  9. OpenDayLight Helium实验一 OpenDaylight的C/S模式实验

    本文基于:OpenDaylight的C/S模式实验而成 C/S 结构,即大家熟知的客户机和服务器结构.它是软件系统体系结构,通过它可以充分利用两端硬件环境的优势,将任务合理分配到Client端和Ser ...

  10. Spring Cloud 开发的一些推荐规划

    1.提供一个统一的 父 pom 依赖    作用:统一版本与引入必要依赖 2.提供一个模板模型. 作用: 开发人员不必关系具体基础启动项 3.提供一个统一基础配置模型 作用: 开发人员不比太过关注与必 ...