最短路径问题，djkstra算法

Problem C: Shortest Path

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 1007  Solved: 348
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Description

Abrao is a beautiful country. It has N cities (1 ≤ N≤ 2,500) and C (1 ≤ C ≤ 10,000) highways to connect adjacent cities. Each city is assigned with a unique number from 1 to N. You are required to calculate the length of the shortest path from one city to another.

Input

The first line contains four integers N, C, S and T. N and C has been defined above. S and T (1 ≤ S,T≤ 2,500) are the numbers of two cities. The information about the highways is defined in the following C lines, each one of which contains three integers v_i,v_j and w  indicating a high way between v_i and v_j with the distance equal to w (there may be multiple highways between two cities). You are required to calculate the length of the shortest path from the city S to T. 

Output

The output is the length of the shortest path from the city S to T.

Sample Input

7 11 5 42 4 21 4 37 2 23 4 35 7 57 3 36 1 16 3 42 4 35 6 37 2 1

Sample Output

7

#include <iostream>#include <vector>#include <limits.h>using namespace std;#define UNVISITED false#define VISITED trueclass Edge{private:int destination;int weight;public:Edge(int a, int b){destination = a;weight = b;}Edge(){destination = (int)INT_MAX;weight = (int)INT_MAX;}int Getdestination(){return destination;}int Getweight(){return weight;}};class graph{private:int numofvertices;int numofhighways;vector<Edge>* graphvertices;bool* mask;public:graph(int numver, int numhigh){numofvertices = numver;numofhighways = numhigh;graphvertices = new vector<Edge>[numver];mask = new bool[numver];for (int i = 0; i < numver; i++){mask[i] = UNVISITED;}}void Insert(int start, int destination, int weight){Edge tmp(destination, weight);graphvertices[start - 1].push_back(tmp);}vector<Edge> Getverticelist(int i){return graphvertices[i];}int Getnumofvertices(){return numofvertices;}int Getnumofhighways(){return numofhighways;}bool Getmask(int i){return mask[i];}void Setmask(int i, bool change){mask[i] = change;}};int Minvertice(graph* G, int* D){int min = INT_MAX;int v = 0;for (int i = 0; i < G->Getnumofvertices(); i++){if (G->Getmask(i) == UNVISITED&&D[i] < min){v = i;min = D[i];}}return v;}void djkstra(graph* G, int* D, int s){for (int i = 0; i < G->Getnumofvertices(); i++){D[i] = INT_MAX;}D[s - 1] = 0;for (int i = 0; i < G->Getnumofvertices(); i++){int v = Minvertice(G, D);G->Setmask(v, VISITED);for (int i = 0; i < G->Getverticelist(v).size(); i++){int w = G->Getverticelist(v).at(i).Getdestination() - 1;if (D[w] >(D[v] + G->Getverticelist(v).at(i).Getweight())){D[w] = (D[v] + G->Getverticelist(v).at(i).Getweight());}}}}int main(){int numcities, numhighway, Start, Terminal;cin >> numcities >> numhighway >> Start >> Terminal;graph* path = new graph(numcities, numhighway);int* D = new int[numcities];while (numhighway--){int start, destination, weight;cin >> start >> destination >> weight;path->Insert(start, destination, weight);path->Insert(destination, start, weight);}djkstra(path, D, Start);cout << D[Terminal - 1] << endl;return 0;}

经验：尽量使用c++库内包含的数据结构，自己写链表还有队列之类的会很崩溃的，而且浪费时间又做不好。

INT_MAX要引入<limits.h>头文件