现在给你一个深圳地铁图。小明从市民中心上车，计算他到深圳所有地铁站所需时间(简化每个站到下一个站只花2分钟)。这就是迪杰斯特拉算法干的事。
历史：Dijkstra thought about the shortest path problem when working at the Mathematical Center in Amsterdam in 1956 as a programmer to demonstrate capabilities of a new computer called ARMAC. His objective was to choose both a problem as well as an answer (that would be produced by computer) that non-computing people could understand. He designed the shortest path algorithm and later implemented it for ARMAC for a slightly simplified transportation map of 64 cities in the Netherlands (64, so that 6 bits would be sufficient to encode the city number).[1] A year later, he came across another problem from hardware engineers working on the institute's next computer: minimize the amount of wire needed to connect the pins on the back panel of the machine. As a solution, he re-discovered the algorithm known as Prim's minimal spanning tree algorithm (known earlier to Jarník, and also rediscovered by Prim).[5][6] Dijkstra published the algorithm in 1959, two years after Prim and 29 years after Jarník.。
大概意思就是D这个人呐在MC工作，他在检验当时一个叫ARMAC机的能力。他想设计一个问题和算法让普通人都能明白，于是拿起了荷兰64座城市地图。之后又做了一些事情，比如PRIM算法，以及利用这些算法解决了插接线的问题啥的。后来他在1959年就把这算法发表了。
代码：
#include <iostream> using namespace std; const int maxnum = 100; const int maxint = 999999; // 各数组都从下标1开始 int dist[maxnum]; // 表示当前点到源点的最短路径长度 int prev[maxnum]; // 记录当前点的前一个结点 int c[maxnum][maxnum]; // 记录图的两点间路径长度 int n, line; // 图的结点数和路径数 // n -- n nodes // v -- the source node // dist[] -- the distance from the ith node to the source node // prev[] -- the previous node of the ith node // c[][] -- every two nodes' distance void Dijkstra(int n, int v, int *dist, int *prev, int c[maxnum][maxnum]) { //步骤1------------初始化------------- bool s[maxnum]; // 判断是否已存入该点到S集合中 for(int i=1; i<=n; ++i) { dist[i] = c[v][i]; s[i] = 0; // 初始都未用过该点 if(dist[i] == maxint) prev[i] = 0; else prev[i] = v; } dist[v] = 0; s[v] = 1; // 依次将未放入S集合的结点中，取dist[]最小值的结点，放入结合S中 // 一旦S包含了所有V中顶点，dist就记录了从源点到所有其他顶点之间的最短路径长度 // 注意是从第二个节点开始，第一个为源点 //-----------步骤2--找到最小的并纳入------------- //maxint 是无穷 //v是0点 //tmp是 找的最小点u的dis for(i=2; i<=n; ++i) { int tmp = maxint; int u = v; for(int j=1; j<=n; ++j) if((!s[j]) && dist[j]<tmp) { u = j; // u保存当前邻接点中距离最小的点的号码 tmp = dist[j]; } s[u] = 1; // 表示u点已存入S集合中 //----------步骤3--更新dist------------------ //u 是刚刚找到最小点 //s 是并入集合 //newdist,dist[u]+c[u][j] 是0-u点距离+u到个点距离 //dist[j]是个0-点的距离 //prev[j]是个点的前一点 for(j=1; j<=n; ++j) if((!s[j]) && c[u][j]<maxint) { int newdist = dist[u] + c[u][j]; if(newdist < dist[j]) { dist[j] = newdist; prev[j] = u; } } } //步骤2,3重复n-1 } // 查找从源点v到终点u的路径，并输出 void searchPath(int *prev,int v, int u) { int que[maxnum]; int tot = 1; que[tot] = u; tot++; int tmp = prev[u]; while(tmp != v) { que[tot] = tmp; tot++; tmp = prev[tmp]; } que[tot] = v; for(int i=tot; i>=1; --i) if(i != 1) cout << que[i] << " -> "; else cout << que[i] << endl; } int main() { freopen("input.txt", "r", stdin); // 各数组都从下标1开始 // 输入结点数 cin >> n; // 输入路径数 cin >> line; int p, q, len; // 输入p, q两点及其路径长度 // 初始化c[][]为maxint for(int i=1; i<=n; ++i) for(int j=1; j<=n; ++j) c[i][j] = maxint; for(i=1; i<=line; ++i) { cin >> p >> q >> len; if(len < c[p][q]) // 有重边 { c[p][q] = len; // p指向q c[q][p] = len; // q指向p，这样表示无向图 } } for(i=1; i<=n; ++i) dist[i] = maxint; for(i=1; i<=n; ++i) { for(int j=1; j<=n; ++j) printf("%8d", c[i][j]); printf(" "); } Dijkstra(n, 1, dist, prev, c); for(int k=1;k<n;k++) { // 最短路径长度 cout <<"源点到顶点"<<k+1<<" 的最短路径长度:" << dist[k+1] <<" "; // 路径 cout <<"源点到顶点"<<k+1<<" 的路径为: "; searchPath(prev, 1, k+1); } }
例子：
算法步骤如下：
1. 初使时令 S={V0},T={其余顶点}，T中顶点对应的距离值
若存在<V0,Vi>，d(V0,Vi)为<V0,Vi>弧上的权值
若不存在<V0,Vi>，d(V0,Vi)为∝
2. 从T中选取一个其距离值为最小的顶点W且不在S中，加入S
|
|
初始化 |
S1纳入2(10最小) |
S1纳入4(30最小) |
S1纳入3(50最小) |
S1纳入5(60最小) |
|
2 |
10 |
10 |
10 |
10 |
10 |
|
3 |
无穷 |
10+50<无穷 60 |
30+20< 60 50 |
50 |
50 |
|
4 |
30 |
10+无穷>30 30 |
30 |
30 |
30 |
|
5 |
100 |
10+无穷>100 100 |
30+60<100 90 |
50+10<90 60 |
60 |
|
S1 |
{1} |
{1,2} |
{1,2,4} |
{1,2,4,3} |
{1,2,4,3,5} |
|
S2 |
{2,3,4,5} |
{3,4,5} |
{3,5} |
{5} |
{} |
|
dis |
|
Dis2=10 |
Dis4=30 |
Dis3=50 |
Dis5=60 |
|
pre |
Pre2=1 Pre4=1 Pre5=1 |
pre3=2 |
Pre3=4 Pre5=4 |
Pre5=3 |
|