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. Mail.Ru Cup 2018 Round 2 Solution

    A. Metro Solved. 题意: 有两条铁轨,都是单向的,一条是从左往右,一条是从右往左,Bob要从第一条轨道的第一个位置出发,Alice的位置处于第s个位置,有火车会行驶在铁轨上,一共有n个 ...

  2. 2018 Multi-University Training Contest 8 Solution

    A - Character Encoding 题意:用m个$0-n-1$的数去构成k,求方案数 思路:当没有0-n-1这个条件是答案为C(k+m-1, m-1),减去有大于的关于n的情况,当有i个n时 ...

  3. Java backup

    待补充 ........ 0:常用头文件(待补充) import java.util.Arrays; import java.util.HashSet; import java.util.TreeSe ...

  4. 高级bash脚本编程(三)

    高级bash脚本编程 知识点 compound 和 comparison -a 逻辑与 exp1 -a exp2 如果表达式 exp1 和 exp2 都为真的话,那么结果为真. -o 逻辑或 exp1 ...

  5. ubuntu16.04解决tensorflow提示未编译使用SSE3、SSE4.1、SSE4.2、AVX、AVX2、FMA的问题【转】

    本文转载自:https://blog.csdn.net/Nicholas_Wong/article/details/70215127 rticle/details/70215127 在我的机器上出现的 ...

  6. Matlab绘图基础——利用axes(坐标系图形对象)绘制重叠图像 及 一图多轴(一幅图绘制多个坐标轴)

    描述 axes在当前窗口中创建一个包含默认属性坐标系 axes('PropertyName',propertyvalue,...)创建坐标系时,同时指定它的一些属性,没有指定的使用DefaultAxe ...

  7. JS浏览器BOM

    浏览器对象模型 (BOM)  BOM的核心是window,而window对象又具有双重角色,它既是通过js访问浏览器窗口的一个接口,又是一个Global(全局)对象.这意味着在网页中定义的任何对象,变 ...

  8. Python学习札记(二十八) 模块1

    参考:模块 NOTE 1.模块:一个.py文件称为一个模块. 2.代码模块化的意义:a.提升程序的可维护性 b.不用重复造轮子 3.避免模块冲突,解决方法:引入了按目录来组织模块的方法,称为包(Pac ...

  9. HDU 6053 TrickGCD(莫比乌斯反演)

    http://acm.hdu.edu.cn/showproblem.php?pid=6053 题意:给出一个A数组,B数组满足Bi<=Ai. 现在要使得这个B数组的GCD值>=2,求共有多 ...

  10. codeforces 55 div2 C.Title 模拟

    C. Title time limit per test 2 seconds memory limit per test 256 megabytes input standard input outp ...