本题不是讲迪杰斯特拉算法的,而是在该算法基础上用优先队列进行优化
首先回顾dijkstra,每次用一个已经更新过的点的最小值去更新周围的点,因此可以用一个优先队列(从小到大顺序)去保存这些已经更新过的,每次从队列里取队头的元素,判断这个位置的值更新过没有,没更新过就更新,更新过就继续找下一个,依次进行,知道队列空了,跳出循环
#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
int g[2510][2510];
int dist[2510];
int v[2510];
int n, m, a, b;
struct p{
int index, dist;
p(int index, int dist){
this->index = index;
this->dist = dist;
}
bool operator > (const p &a)const{
return this->dist > a.dist;
}
bool operator < (const p &a)const{
return this->dist < a.dist;
}
};
int main()
{
memset(g, 0x3f, sizeof g);
memset(v, 0, sizeof v);
memset(dist, 0x3f, sizeof dist);
scanf("%d %d %d %d", &n, &m, &a, &b);
priority_queue<p, vector<p>, greater<p> > q;
for (int i = 1; i <= m; i++){
int x, y, z;
scanf("%d %d %d", &x, &y, &z);
g[x][y] = z;
g[y][x] = z;
}
for (int i = 1; i <= n; i++){
if(g[a][i] != 0x3f3f3f3f){
dist[i] = g[a][i];
q.push(p(i, dist[i]));
}
}
v[a] = 1;
for(;;){
int ind, mn;
p t(-1,-1);
while(!q.empty() && v[q.top().index] == 1){
q.pop();
}
if(!q.empty()){
t = q.top();
q.pop();
}
ind = t.index;
mn = t.dist;
if(ind == -1)break;
for (int i = 1; i <= n; i++){
if(dist[i] > mn + g[ind][i]){
dist[i] = mn + g[ind][i];
q.push(p(i, dist[i]));
}
}
}
printf("%d", dist[b]);
return 0;
}