Dijkstra算法java实现
2015-06-04 14:52
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看到网上的Dijkstra算法写的都好复杂,我自己就简单的写了一个,例子为书上的例子。
public class Dijkstra {
public static void main(String[] args) {
int arcs[][] = {
{Integer.MAX_VALUE,Integer.MAX_VALUE,10,Integer.MAX_VALUE,30,100},
{Integer.MAX_VALUE,Integer.MAX_VALUE,5,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE},
{Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,50,Integer.MAX_VALUE,Integer.MAX_VALUE},
{Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,10},
{Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,20,Integer.MAX_VALUE,60},
{Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE}
};
//存放的为顶点所在数组的ID和顶点名称
Map<Integer, String> v = new HashMap<Integer,String>();
//不存放源点
//v.put(0,"v0");
v.put(1,"v1");
v.put(2,"v2");
v.put(3,"v3");
v.put(4,"v4");
v.put(5,"v5");
Set<Integer> s = new HashSet<Integer>();
//s中存放尚未找到最短路径的顶点的id
s.addAll(v.keySet());
int minPath;
int id;
int removeId;
Iterator<Integer> it = null;
for(int i=0; i<v.size(); i++) {
removeId = 0;
minPath = Integer.MAX_VALUE;
it = s.iterator();
while(it.hasNext()) {
id = it.next();
//源点到所有顶点的最短距离都存储在arcs[0]数组中
if(minPath>arcs[0][id]) {
minPath = arcs[0][id];
removeId = id;
}
}
if(minPath == Integer.MAX_VALUE) {
System.out.println("源点v0到其余顶点的距离都为不可达");
break;
}
s.remove(removeId);
System.out.println("源点v0到顶点" + v.get(removeId) + "的最短距离为:" + arcs[0][removeId]);
it = s.iterator();
while(it.hasNext()) {
id = it.next();
if(arcs[removeId][id] == Integer.MAX_VALUE) continue;
if(arcs[0][id] > minPath + arcs[removeId][id]) {
arcs[0][id] = minPath + arcs[removeId][id];
}
}
}
}
}
public class Dijkstra {
public static void main(String[] args) {
int arcs[][] = {
{Integer.MAX_VALUE,Integer.MAX_VALUE,10,Integer.MAX_VALUE,30,100},
{Integer.MAX_VALUE,Integer.MAX_VALUE,5,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE},
{Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,50,Integer.MAX_VALUE,Integer.MAX_VALUE},
{Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,10},
{Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,20,Integer.MAX_VALUE,60},
{Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE}
};
//存放的为顶点所在数组的ID和顶点名称
Map<Integer, String> v = new HashMap<Integer,String>();
//不存放源点
//v.put(0,"v0");
v.put(1,"v1");
v.put(2,"v2");
v.put(3,"v3");
v.put(4,"v4");
v.put(5,"v5");
Set<Integer> s = new HashSet<Integer>();
//s中存放尚未找到最短路径的顶点的id
s.addAll(v.keySet());
int minPath;
int id;
int removeId;
Iterator<Integer> it = null;
for(int i=0; i<v.size(); i++) {
removeId = 0;
minPath = Integer.MAX_VALUE;
it = s.iterator();
while(it.hasNext()) {
id = it.next();
//源点到所有顶点的最短距离都存储在arcs[0]数组中
if(minPath>arcs[0][id]) {
minPath = arcs[0][id];
removeId = id;
}
}
if(minPath == Integer.MAX_VALUE) {
System.out.println("源点v0到其余顶点的距离都为不可达");
break;
}
s.remove(removeId);
System.out.println("源点v0到顶点" + v.get(removeId) + "的最短距离为:" + arcs[0][removeId]);
it = s.iterator();
while(it.hasNext()) {
id = it.next();
if(arcs[removeId][id] == Integer.MAX_VALUE) continue;
if(arcs[0][id] > minPath + arcs[removeId][id]) {
arcs[0][id] = minPath + arcs[removeId][id];
}
}
}
}
}
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