tsp的动态规划实现，最短长度和最短路径

来源：互联网发布：python win32api 函数编辑：程序博客网时间：2024/05/13 22:41

#include "stdio.h"#include "string.h"#define INF 999999999typedef struct _node {int value;      //indicate the value of the minimum distance of the state in this phase.int min_index;  //keep trace of the index which is just before this node}node;node table[40][6];int bit[6];int n = 6;int path[5];int dis[6][6] = {0, 10, 20, 30, 40, 50,12, 0 ,18, 30, 25, 21,23, 19, 0, 5,  10, 15,34, 32, 4, 0,  8,  16,45, 27, 11,10, 0,  18,56, 22, 16,20, 12,  0};int  init(){int i;int j;int t;memset( table, -1, sizeof(table) );for( i = 0; i < n; i++ ){bit[i] = 1<<i;}t = 1;for( i = 1; i < n; i++ ){table[ t<<( i - 1 ) ][i].value = dis[0][i];}return 1;}int getTable(int s,int k){int t,tt;int i;int min = INF;if(table[s][k].value != -1){return table[s][k].value;}t = s & (~bit[k - 1]);for(i = 1; i < n; i++){tt=t & bit[i-1];if(tt > 0){if(getTable(t, i) + dis[i][k] < min){min = getTable(t, i) + dis[i][k];table[s][k].min_index = i;}}}table[s][k].value = min;return table[s][k].value;}int main(){int i;int t;int ret;t=0;int set;ret=INF;init();for(i = 1; i < n; i++){t = t<<1;t = t|1;}set = t;for(i = 1; i < n; i++){if(getTable(t,i)+dis[i][0]<ret){ret=getTable(t,i)+dis[i][0];path[4] = i;}}  for (int i = 4;i > 0;i--){int index = table[set][path[i]].min_index;path[i - 1] = index;set = set & (~bit[path[i] - 1]);}    printf("the intermediate vertice are ：\n");for(int i = 0; i < 5;++i){printf("%d ",path[i] + 1);}printf("\nthe minimum distance is ：\n%d\n",ret);getchar();return 1;}

主体代码转自http://blog.sina.com.cn/s/blog_48f85e1d0100qz58.html
记c(s,k)状态为从v0，遍历了S中所有的节点，最后停在K点的所有路径的最小值,那么状态转移方程为：
c(s,k)=min(c(s-{vk},vi)+g[vi][k])~~~~~~vi属于s-{vk}
初始状态很明显，就是c(ti,i)=g[0][i],ti表示只含有v0一个元素的集合.

具体实现是用二进制位来表示节点是否在集合中