要找出最短路径,其实就是从起点遍历所有能到达的顶点,然后计算他们的权重。Dijkstra算法核心在于边的松弛(relax),可以想象成一根绷紧的橡皮筋,让它放松下来。即是计算源点(s)经过当前点(v)到目标点(w)的权重,如果比目标点(w)之前的权重要小,就替换掉。最终的结果就是生成一颗最小路径树。这个算法和prim算法非常相似,甚至就是prim即时算法的变种。如果加权无向图和加权有向图的边和权重对应,最短路径树和最小生成树其实是等价的。
Dijkstra算法并不能处理权重为负数的边。
例如,D[3] = 2表示从起始点到顶点3的路径相对最小长度为2。这里强调相对就是说在算法执行过程中D的值是在不断逼近最终结果但在过程中不一定就等于长度。
2.D的初始状态为:若从
到
有弧(即从
到
存在连接边),则D
为弧上的权值(即为从
到
的边的权值);否则置D
为∞。
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=efa941b69c3df8dca23d83a0cd3f23/29381f30e924b899366736796d061d950a7bf61c.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D12/sign=8da8de7e48fbfbedd859327d78f0572b/242dd42a2834349bd43c21c4caea15ce36d3bebd.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=efa941b69c3df8dca23d83a0cd3f23/29381f30e924b899366736796d061d950a7bf61c.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D12/sign=8da8de7e48fbfbedd859327d78f0572b/242dd42a2834349bd43c21c4caea15ce36d3bebd.jpg)
![](https://gss0.bdstatic.com/-4o3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D17/sign=480315771e950a7b71354ac30ad18932/7aec54e736d12f2ebd33e0214cc2d562853568b8.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=efa941b69c3df8dca23d83a0cd3f23/29381f30e924b899366736796d061d950a7bf61c.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D12/sign=8da8de7e48fbfbedd859327d78f0572b/242dd42a2834349bd43c21c4caea15ce36d3bebd.jpg)
![](https://gss0.bdstatic.com/-4o3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D17/sign=480315771e950a7b71354ac30ad18932/7aec54e736d12f2ebd33e0214cc2d562853568b8.jpg)
显然,长度为 D
= Min{ D |
∈V } 的路径就是从
出发到顶点
的长度最短的一条路径,此路径为(
)。
![](https://gss3.bdstatic.com/-Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D18/sign=227e4f8b818ba61edbeecc27403419f1/4bed2e738bd4b31c144addfa84d6277f9e2ff879.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D12/sign=8da8de7e48fbfbedd859327d78f0572b/242dd42a2834349bd43c21c4caea15ce36d3bebd.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=efa941b69c3df8dca23d83a0cd3f23/29381f30e924b899366736796d061d950a7bf61c.jpg)
![](https://gss3.bdstatic.com/-Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D16/sign=eef798ace41190ef05fb96d9cf1b2064/43a7d933c895d143c689ac6270f082025baf07f7.jpg)
![](https://gss0.bdstatic.com/94o3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D28/sign=8922c04fb23533faf1b69426a9d31b22/8718367adab44aedbe89f7a5b01c8701a08bfba8.jpg)
3.那么,下一条长度次短的是哪一条呢?也就是找到从源点
到下一个顶点的最短路径长度所对应的顶点,且这条最短路径长度仅次于从源点
到顶点
的最短路径长度。
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=efa941b69c3df8dca23d83a0cd3f23/29381f30e924b899366736796d061d950a7bf61c.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=efa941b69c3df8dca23d83a0cd3f23/29381f30e924b899366736796d061d950a7bf61c.jpg)
![](https://gss3.bdstatic.com/-Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D16/sign=eef798ace41190ef05fb96d9cf1b2064/43a7d933c895d143c689ac6270f082025baf07f7.jpg)
假设该次短路径的终点是
,则可想而知,这条路径要么是(
),或者是(
)。它的长度或者是从
到
的弧上的权值,或者是D
加上从
到
的弧上的权值。
![](https://gss0.bdstatic.com/94o3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D14/sign=12efb060718b4710ca2ff9c8c2ce63f2/63d9f2d3572c11df01a2f762602762d0f703c204.jpg)
![](https://gss1.bdstatic.com/-vo3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D29/sign=174cae78b0de9c82a265fe866c81033f/4b90f603738da977bc06cc6ab351f8198618e342.jpg)
![](https://gss1.bdstatic.com/-vo3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D49/sign=f00c828bac51f3dec7b2b86d95ee39c1/ac4bd11373f0820237b5db7e48fbfbedab641b4f.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=efa941b69c3df8dca23d83a0cd3f23/29381f30e924b899366736796d061d950a7bf61c.jpg)
![](https://gss0.bdstatic.com/94o3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D14/sign=12efb060718b4710ca2ff9c8c2ce63f2/63d9f2d3572c11df01a2f762602762d0f703c204.jpg)
![](https://gss3.bdstatic.com/-Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D18/sign=227e4f8b818ba61edbeecc27403419f1/4bed2e738bd4b31c144addfa84d6277f9e2ff879.jpg)
![](https://gss3.bdstatic.com/-Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D16/sign=5833cf20fb1986184547eb824bedbecd/902397dda144ad34b2aacdc6d3a20cf431ad855e.jpg)
![](https://gss0.bdstatic.com/94o3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D14/sign=12efb060718b4710ca2ff9c8c2ce63f2/63d9f2d3572c11df01a2f762602762d0f703c204.jpg)
4.一般情况下,假设S为已求得的从源点
出发的最短路径长度的顶点的集合,则可证明:下一条次最短路径(设其终点为
)要么是弧(
),或者是从源点
出发的中间只经过S中的顶点而最后到达顶点
的路径。
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=efa941b69c3df8dca23d83a0cd3f23/29381f30e924b899366736796d061d950a7bf61c.jpg)
![](https://gss1.bdstatic.com/-vo3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=6605e2a1f403918fd3d131fa51c3e7/4ec2d5628535e5dde233109e75c6a7efce1b6257.jpg)
![](https://gss3.bdstatic.com/-Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D25/sign=583f7537f11fbe09185ec4116a60e51b/dc54564e9258d109902ae3f3d258ccbf6d814da1.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=efa941b69c3df8dca23d83a0cd3f23/29381f30e924b899366736796d061d950a7bf61c.jpg)
![](https://gss1.bdstatic.com/-vo3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=6605e2a1f403918fd3d131fa51c3e7/4ec2d5628535e5dde233109e75c6a7efce1b6257.jpg)
因此,下一条长度次短的的最短路径长度必是D
= Min{ D
|
∈V-S },其中D
要么是弧(
)上的权值,或者是D
(
∈S)和弧(
,
)上的权值之和。
![](https://gss3.bdstatic.com/-Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D18/sign=227e4f8b818ba61edbeecc27403419f1/4bed2e738bd4b31c144addfa84d6277f9e2ff879.jpg)
![](https://gss0.bdstatic.com/-4o3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D17/sign=480315771e950a7b71354ac30ad18932/7aec54e736d12f2ebd33e0214cc2d562853568b8.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D12/sign=8da8de7e48fbfbedd859327d78f0572b/242dd42a2834349bd43c21c4caea15ce36d3bebd.jpg)
![](https://gss0.bdstatic.com/-4o3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D17/sign=480315771e950a7b71354ac30ad18932/7aec54e736d12f2ebd33e0214cc2d562853568b8.jpg)
![](https://gss2.bdstatic.com/9fo3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D27/sign=992a7b9ca864034f0bcdc501aec35574/d52a2834349b033b8964499b16ce36d3d439bd87.jpg)
![](https://gss3.bdstatic.com/-Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D17/sign=037e5132ae4bd11300cdb3355baf5d12/eaf81a4c510fd9f91cc1e028262dd42a2934a499.jpg)
![](https://gss0.bdstatic.com/94o3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D14/sign=12efb060718b4710ca2ff9c8c2ce63f2/63d9f2d3572c11df01a2f762602762d0f703c204.jpg)
![](https://gss0.bdstatic.com/94o3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D14/sign=12efb060718b4710ca2ff9c8c2ce63f2/63d9f2d3572c11df01a2f762602762d0f703c204.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D12/sign=8da8de7e48fbfbedd859327d78f0572b/242dd42a2834349bd43c21c4caea15ce36d3bebd.jpg)
算法描述如下:
1)令arcs表示弧上的权值。若弧不存在,则置arcs为∞(在本程序中为MAXCOST)。S为已找到的从
出发的的终点的集合,初始状态为空集。那么,从
出发到图上其余各顶点
可能达到的长度的初值为D=arcs[Locate Vex(G,
)],
∈V;
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=efa941b69c3df8dca23d83a0cd3f23/29381f30e924b899366736796d061d950a7bf61c.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=efa941b69c3df8dca23d83a0cd3f23/29381f30e924b899366736796d061d950a7bf61c.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D12/sign=8da8de7e48fbfbedd859327d78f0572b/242dd42a2834349bd43c21c4caea15ce36d3bebd.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D12/sign=8da8de7e48fbfbedd859327d78f0572b/242dd42a2834349bd43c21c4caea15ce36d3bebd.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D12/sign=8da8de7e48fbfbedd859327d78f0572b/242dd42a2834349bd43c21c4caea15ce36d3bebd.jpg)
2)选择
,使得D
=Min{ D |
∈V-S } ;
![](https://gss3.bdstatic.com/-Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D16/sign=eef798ace41190ef05fb96d9cf1b2064/43a7d933c895d143c689ac6270f082025baf07f7.jpg)
![](https://gss3.bdstatic.com/-Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D18/sign=227e4f8b818ba61edbeecc27403419f1/4bed2e738bd4b31c144addfa84d6277f9e2ff879.jpg)
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D12/sign=8da8de7e48fbfbedd859327d78f0572b/242dd42a2834349bd43c21c4caea15ce36d3bebd.jpg)
3)修改从
出发的到集合V-S中任一顶点
的最短路径长度。
![](https://gss3.bdstatic.com/7Po3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D9/sign=efa941b69c3df8dca23d83a0cd3f23/29381f30e924b899366736796d061d950a7bf61c.jpg)
![](https://gss0.bdstatic.com/94o3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D14/sign=12efb060718b4710ca2ff9c8c2ce63f2/63d9f2d3572c11df01a2f762602762d0f703c204.jpg)
按路径长度递增次序产生算法:
把顶点集合V分成两组:
(1)S:已求出的顶点的集合(初始时只含有源点V0)
(2)V-S=T:尚未确定的顶点集合
将T中顶点按递增的次序加入到S中,保证:
(1)从源点V0到S中其他各顶点的长度都不大于从V0到T中任何顶点的最短路径长度
(2)每个顶点对应一个距离值
S中顶点:从V0到此顶点的长度
T中顶点:从V0到此顶点的只包括S中顶点作中间顶点的最短路径长度
依据:可以证明V0到T中顶点Vk的,或是从V0到Vk的直接路径的权值;或是从V0经S中顶点到Vk的路径权值之和
(反证法可证)
求最短路径步骤
算法步骤如下:
G={V,E}
1. 初始时令 S={V0},T=V-S={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中选取一个与S中顶点有关联边且权值最小的顶点W,加入到S中
3. 对其余T中顶点的距离值进行修改:若加进W作中间顶点,从V0到Vi的距离值缩短,则修改此距离值
重复上述步骤2、3,直到S中包含所有顶点,即W=Vi为止