多源最短路径Floyd、Floyd求最小环【模板】

Floyd算法：用来找出每对点之间的最短距离。图可以是无向图，也可以是有向图，边权可为正，也可以为负，唯一要求是不能有负环。
1.初始化：将Map[][]中的数据复制到Dist[][]中作为每对顶点之间的最短路径的初值，Pre[i][j] = i 表示 i 到 j 路径中 j 的前一节点。
2. k 从 1 到 N 循环 N 次，每次循环中，枚举图中不同的两点 i，j，如果Dist[i][j] > Dist[i][k] + Dist[k][j]，则更新Dist[i][j] = Dist[i][k] + Dist[k][j]，更新Pre[i][j] = Pre[k][j]。
只要图中不存在负环就可以得出正确的答案，关于Floyd算法对负环的判定，参考下边Floyd求最小环。

<code class="hljs markdown has-numbering" style="display: block; padding: 0px; color: inherit; box-sizing: border-box; font-family: 'Source Code Pro', monospace;font-size:undefined; white-space: pre; border-radius: 0px; word-wrap: normal; background: transparent;">const int MAXN = 110;
const int INF = 0xffffff0;int Map[<span class="hljs-link_label" style="box-sizing: border-box;">MAXN</span>][<span class="hljs-link_reference" style="box-sizing: border-box;">MAXN</span>], Dist[<span class="hljs-link_label" style="box-sizing: border-box;">MAXN</span>][<span class="hljs-link_reference" style="box-sizing: border-box;">MAXN</span>],Pre[<span class="hljs-link_label" style="box-sizing: border-box;">MAXN</span>][<span class="hljs-link_reference" style="box-sizing: border-box;">MAXN</span>];
//Pre[<span class="hljs-link_label" style="box-sizing: border-box;">i</span>][<span class="hljs-link_reference" style="box-sizing: border-box;">j</span>] = i表示i到j路径中j的前一节点
void Floyd(int N)
{   //初始化
<span class="hljs-code" style="box-sizing: border-box;">    for(int i = 1; i <= N; ++i)</span>
<span class="hljs-code" style="box-sizing: border-box;">    {</span>
<span class="hljs-code" style="box-sizing: border-box;">        for(int j = 1; j <= N; ++j)</span>
<span class="hljs-code" style="box-sizing: border-box;">        {</span>
<span class="hljs-code" style="box-sizing: border-box;">            Dist[i][j] = Map[i][j];</span>
<span class="hljs-code" style="box-sizing: border-box;">            Pre[i][j] = i;</span>
<span class="hljs-code" style="box-sizing: border-box;">        }</span><span class="hljs-code" style="box-sizing: border-box;">    }</span>
<span class="hljs-code" style="box-sizing: border-box;">    for(int k = 1; k <= N; ++k)</span>
<span class="hljs-code" style="box-sizing: border-box;">    {</span>
<span class="hljs-code" style="box-sizing: border-box;">        for(int i = 1; i <= N; ++i)</span>
<span class="hljs-code" style="box-sizing: border-box;">        {</span>
<span class="hljs-code" style="box-sizing: border-box;">            for(int j = 1; j <= N; ++j)</span>
<span class="hljs-code" style="box-sizing: border-box;">            {   //如果Dist[i][j] > Dist[i][k] + Dist[k][j]，则更新</span>
<span class="hljs-code" style="box-sizing: border-box;">                if(Dist[i][k] != INF && Dist[k][j] != INF && Dist[i][k] + Dist[k][j] < Dist[i][j])</span>
<span class="hljs-code" style="box-sizing: border-box;">                {</span>
<span class="hljs-code" style="box-sizing: border-box;">                    Dist[i][j] = Dist[i][k] + Dist[k][j];</span>
<span class="hljs-code" style="box-sizing: border-box;">                    Pre[i][j] = Pre[k][j];  //更新Pre[i][j]</span>
<span class="hljs-code" style="box-sizing: border-box;">                }    </span>
<span class="hljs-code" style="box-sizing: border-box;">            }</span>
<span class="hljs-code" style="box-sizing: border-box;">        }</span>
<span class="hljs-code" style="box-sizing: border-box;">    }</span>
}</code><ul class="pre-numbering" style="box-sizing: border-box; position: absolute; width: 50px; top: 0px; left: 0px; margin: 0px; padding: 6px 0px 40px; border-right-width: 1px; border-right-style: solid; border-right-color: rgb(221, 221, 221); list-style: none; text-align: right; background-color: rgb(238, 238, 238);"><li style="box-sizing: border-box; padding: 0px 5px;">1</li><li style="box-sizing: border-box; padding: 0px 5px;">2</li><li style="box-sizing: border-box; padding: 0px 5px;">3</li><li style="box-sizing: border-box; padding: 0px 5px;">4</li><li style="box-sizing: border-box; padding: 0px 5px;">5</li><li style="box-sizing: border-box; padding: 0px 5px;">6</li><li style="box-sizing: border-box; padding: 0px 5px;">7</li><li style="box-sizing: border-box; padding: 0px 5px;">8</li><li style="box-sizing: border-box; padding: 0px 5px;">9</li><li style="box-sizing: border-box; padding: 0px 5px;">10</li><li style="box-sizing: border-box; padding: 0px 5px;">11</li><li style="box-sizing: border-box; padding: 0px 5px;">12</li><li style="box-sizing: border-box; padding: 0px 5px;">13</li><li style="box-sizing: border-box; padding: 0px 5px;">14</li><li style="box-sizing: border-box; padding: 0px 5px;">15</li><li style="box-sizing: border-box; padding: 0px 5px;">16</li><li style="box-sizing: border-box; padding: 0px 5px;">17</li><li style="box-sizing: border-box; padding: 0px 5px;">18</li><li style="box-sizing: border-box; padding: 0px 5px;">19</li><li style="box-sizing: border-box; padding: 0px 5px;">20</li><li style="box-sizing: border-box; padding: 0px 5px;">21</li><li style="box-sizing: border-box; padding: 0px 5px;">22</li><li style="box-sizing: border-box; padding: 0px 5px;">23</li><li style="box-sizing: border-box; padding: 0px 5px;">24</li><li style="box-sizing: border-box; padding: 0px 5px;">25</li><li style="box-sizing: border-box; padding: 0px 5px;">26</li><li style="box-sizing: border-box; padding: 0px 5px;">27</li><li style="box-sizing: border-box; padding: 0px 5px;">28</li><li style="box-sizing: border-box; padding: 0px 5px;">29</li><li style="box-sizing: border-box; padding: 0px 5px;">30</li><li style="box-sizing: border-box; padding: 0px 5px;">31</li></ul>

如果求点u到点v能达到的最长边尽可能短的路径上最长边为多少，将循环内部改为如下代码：

<code class="hljs cs has-numbering" style="display: block; padding: 0px; color: inherit; box-sizing: border-box; font-family: 'Source Code Pro', monospace;font-size:undefined; white-space: pre; border-radius: 0px; word-wrap: normal; background: transparent;"><span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> tMax;   <span class="hljs-comment" style="color: rgb(136, 0, 0); box-sizing: border-box;">//这里边求的是能达到的路径上最长边最小为多少  </span><span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">if</span>(Dist[i][k] > Dist[k][j])  tMax = Dist[i][k];  <span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">else</span>  tMax = Dist[k][j];  <span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">if</span>(Dist[i][j] > tMax)  Dist[i][j] = tMax;  </code><ul class="pre-numbering" style="box-sizing: border-box; position: absolute; width: 50px; top: 0px; left: 0px; margin: 0px; padding: 6px 0px 40px; border-right-width: 1px; border-right-style: solid; border-right-color: rgb(221, 221, 221); list-style: none; text-align: right; background-color: rgb(238, 238, 238);"><li style="box-sizing: border-box; padding: 0px 5px;">1</li><li style="box-sizing: border-box; padding: 0px 5px;">2</li><li style="box-sizing: border-box; padding: 0px 5px;">3</li><li style="box-sizing: border-box; padding: 0px 5px;">4</li><li style="box-sizing: border-box; padding: 0px 5px;">5</li><li style="box-sizing: border-box; padding: 0px 5px;">6</li><li style="box-sizing: border-box; padding: 0px 5px;">7</li></ul>

Floyd求最小环
不能在Map[][]数组上直接计算，因为判断过程中用到了Map[][]原始值。

<code class="hljs cpp has-numbering" style="display: block; padding: 0px; color: inherit; box-sizing: border-box; font-family: 'Source Code Pro', monospace;font-size:undefined; white-space: pre; border-radius: 0px; word-wrap: normal; background: transparent;"><span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">const</span> <span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> MAXN = <span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">110</span>;
<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">const</span> <span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> INF = <span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">0xffffff0</span>;
<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> temp,Map[MAXN][MAXN],Dist[MAXN][MAXN],pre[MAXN][MAXN],ans[MAXN*<span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">3</span>];<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">void</span> Solve(<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> i,<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> j,<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> k)
{temp = <span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">0</span>;   <span class="hljs-comment" style="color: rgb(136, 0, 0); box-sizing: border-box;">//回溯，存储最小环</span><span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">while</span>(i != j){ans[temp++] = j;j = pre[i][j];}ans[temp++] = i;ans[temp++] = k;
}
<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">void</span> Floyd(<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> N)
{<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">for</span>(<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> i = <span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">1</span>; i <= N; ++i)<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">for</span>(<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> j = <span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">1</span>; j <= N; ++j){Dist[i][j] = Map[i][j];pre[i][j] = i;}<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> MinCircle = INF;    <span class="hljs-comment" style="color: rgb(136, 0, 0); box-sizing: border-box;">//最小环</span><span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">for</span>(<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> k = <span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">1</span>; k <= N; ++k){<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">for</span>(<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> i = <span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">1</span>; i <= N; ++i){<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">for</span>(<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> j = <span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">1</span>; j <= N; ++j){<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">if</span>(i != j && Dist[i][j] != INF && Map[i][k] != INF && Map[k][j] != INF&& Dist[i][j] + Map[i][k] + Map[k][j] < MinCircle){MinCircle = min(MinCircle, Dist[i][j] + Map[i][k] + Map[k][j]);Solve(i,j,k);   <span class="hljs-comment" style="color: rgb(136, 0, 0); box-sizing: border-box;">//回溯存储最小环</span>}}}<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">for</span>(<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> i = <span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">1</span>; i <= N; ++i){<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">for</span>(<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> j = <span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">1</span>; j <= N; ++j){<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">if</span>(Dist[i][k] != INF && Dist[k][j] != INF &&Dist[i][k] + Dist[k][j] < Dist[i][j]){Dist[i][j] = Dist[i][k] + Dist[k][j];pre[i][j] = pre[k][j];  <span class="hljs-comment" style="color: rgb(136, 0, 0); box-sizing: border-box;">//记录点i到点j的路径上，j前边的点</span>}}}}<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">if</span>(MinCircle == INF)    <span class="hljs-comment" style="color: rgb(136, 0, 0); box-sizing: border-box;">//不存在环</span>{<span class="hljs-built_in" style="color: rgb(102, 0, 102); box-sizing: border-box;">printf</span>(<span class="hljs-string" style="color: rgb(0, 136, 0); box-sizing: border-box;">"No solution.\n"</span>);<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">return</span>;}<span class="hljs-comment" style="color: rgb(136, 0, 0); box-sizing: border-box;">//如果求出最小环为负的，原图必定存在负环</span><span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">for</span>(<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">int</span> i = <span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">0</span>;i < temp; ++i)    <span class="hljs-comment" style="color: rgb(136, 0, 0); box-sizing: border-box;">//输出最小环</span><span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">if</span>(i != temp-<span class="hljs-number" style="color: rgb(0, 102, 102); box-sizing: border-box;">1</span>)<span class="hljs-built_in" style="color: rgb(102, 0, 102); box-sizing: border-box;">printf</span>(<span class="hljs-string" style="color: rgb(0, 136, 0); box-sizing: border-box;">"%d "</span>,ans[i]);<span class="hljs-keyword" style="color: rgb(0, 0, 136); box-sizing: border-box;">else</span><span class="hljs-built_in" style="color: rgb(102, 0, 102); box-sizing: border-box;">printf</span>(<span class="hljs-string" style="color: rgb(0, 136, 0); box-sizing: border-box;">"%d\n"</span>,ans[i]);
}</code>

多源最短路径Floyd、Floyd求最小环【模板】相关推荐

最短路径：Dijkstra算法（求单源最短路径）Floyd算法（求各顶点之间最短路径）
最短路径: 在一个带权图中,顶点V0到图中任意一个顶点Vi的一条路径所经过边上的权值之和,定义为该路径的带权路径长度,把带权路径最短的那条路径称为最短路径. DiskStra算法: 求单源最短路径,即 ...
短小精悍的多源最短路径算法—Floyd算法
前言在图论中,在寻路最短路径中除了Dijkstra算法以外,还有Floyd算法也是非常经典,然而两种算法还是有区别的,Floyd主要计算多源最短路径. 在单源正权值最短路径,我们会用Dijkstra ...
floyd求最小环模板
http://www.cnblogs.com/Yz81128/archive/2012/08/15/2640940.html 求最小环 floyd求最小环 2011-08-14 9:42 1 定义: ...
Floyd算法求最小环
/**算法引入:*求一个图G中的最小环路的朴素算法为:每次找到一条边,删除了求这两点之间的最短路径;*若能求出,则这条最短路径与原来的边构成一个环,不过时间复杂度略高;**算法思想;*Floyd算法是 ...
图论-全源最短路径-对比Floyd算法与暴力Dijkstra算法
题目输入顶点数N,有向边数M,接下来M行输入格式为u,v,w分别代表两个顶点u,v和两点之间边的权值w.输出全源最短路径输入样例: 6 8 0 1 1 0 3 4 0 4 4 1 3 2 2 5 ...
HDU1599 find the mincost route Floyd算法求最小环
find the mincost route 杭州有N个景区,景区之间有一些双向的路来连接,现在8600想找一条旅游路线,这个路线从A点出发并且最后回到A点,假设经过的路线为V1,V2,....VK, ...
图的单源最短路径，Floyd算法（数据结构c++）
这个算法结构很是简单,但是理解还是有一定的困难,一开始做的时候想不明白,跟着算法自己动手画画就知道这个算法具体是怎么回事了. 时间复杂度是O(N*3) 算法有点动态规划的意思,有两个数组,一个(dis ...
poj 1734 Sightseeing trip(floyd 拓展求最小环）
对于floyd 拓展为什么可以判断出最小环还是不是太理解苟且先贴下模版吧感谢:心_随_梦_翔的分析 http://hi.baidu.com/%D0%C4_%CB%E6_%C3%CE_%CF% ...
图论-单源最短路径算法（拓扑，Dijkstra，Floyd，SPFA）
前言单源最短路径是学习图论算法的入门级台阶,但刚开始看的时候就蒙了,什么有环没环,有负权没负权,下面就来总结一下求单源最短路径的所有算法以及其适用的情况. 单源最短路径设定图中一个点为源点,求其他 ...
hdu 1599(Floyd求最小环)
find the mincost route Time Limit: 1000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/O ...

多源最短路径Floyd、Floyd求最小环【模板】

多源最短路径Floyd、Floyd求最小环【模板】相关推荐

最新文章

热门文章