传教士与野人过河问题：
任意时刻，左岸、右岸、船上如果传教士人数少于野人人数，传教士就会被野人吃掉。当然野人会划船。传教士人数为0也是可以的。
启发函数 f=g+h. g当前结点所在解空间树的深度。h=m+c-2*b. m,c分别是当前状态下左岸传教士和野人的数目。b=1表示当前船在左岸停靠。b=0表示当前状态船在右岸。
#include<vector>
#include<algorithm>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
#define MAXM 5
#define MAXC 5
typedef struct Node {
int M, C, B;
int g, h, f;
int parent;//记录父节点在tree中的下标
}Node;
vector<Node> open;
vector<Node> close;
vector<Node> tree;
Node boat[8] = {
{ 0,1,1,0,0,0,-1 },
{ 0,2,1,0,0,0,-1 },
{ 0,3,1,0,0,0,-1 },
{ 1,0,1,0,0,0,-1 },
{ 1,1,1,0,0,0,-1 },
{ 2,0,1,0,0,0,-1 },
{ 2,1,1,0,0,0,-1 },
{ 3,0,1,0,0,0,-1 }
};
void show_open() {//显示open表
vector<Node>::iterator ite;
ite = open.end()-1;
for (; ite >=open.begin(); ite--) {
printf("(%d,%d,%d,%d,%d,%d)
", (*ite).M, (*ite).C, (*ite).B, (*ite).g, (*ite).h, (*ite).f);
if (ite == open.begin()) break;
}
}
void show_close() {//显示close表
vector<Node>::iterator ite;
ite = close.begin();
for (; ite < close.end(); ite++) {
printf("(%d,%d,%d,%d,%d,%d)
", (*ite).M, (*ite).C, (*ite).B, (*ite).g, (*ite).h, (*ite).f);
}
}
bool operator ==(Node a, Node b) {
return(a.M == b.M&&a.C == b.C&&a.B == b.B);
}
bool exit_open(Node p) {//判断节点是否存在open表中
if(find(open.begin(), open.end(), p)==open.end()) return false;
else return true;
}
bool exit_close(Node p) {//判断节点是否存在close表中
if (find(close.begin(), close.end(), p) == close.end()) return false;
else return true;
}
bool comp(Node a, Node b) {
return a.f > b.f;//根据估价函数值降序排列
}
void add_open(Node p) {//open表添加
open.push_back(p);
sort(open.begin(), open.end(), comp);//默认升序排列，这里comp按降序排列
}
void add_close(Node p) {//close表添加
close.push_back(p);
}
void out_open() {//open表删除
open.pop_back();
}
bool judge_Node(Node p) {//判断状态p是否合法
if (p.M > MAXM || p.C > MAXC || p.M < 0 || p.C < 0)//不在范围内。不合法
return false;
/*if (((p.M >= p.C) && (MAXM - p.M >= MAXC - p.C)) || (p.M == MAXM) || (p.M == 0))
return true;
return false;*/
else if (p.M != 0 && p.M < p.C)//左岸传教士人数不为0 并且小于野人数
return false;
else if (MAXM - p.M != 0 && MAXM - p.M < MAXC - p.C)//右岸传教士人数不为0就算了 居然比野人数少！！当然不行
return false;
else return true;
}
void expand(Node p) {//对节点p进行扩展
Node q;
printf("------------------------------------------------------------------------------------
");
printf(" 对结点: ");
printf("( %d %d %d )进行扩展
", p.M, p.C, p.B);
for (int i = 0; i < 8; i++) {
if (p.B == 1) {//p船在左岸
q.M = p.M - boat[i].M;
q.C = p.C - boat[i].C;
q.B = p.B - boat[i].B;
}
else {//p船在右岸
q.M = p.M + boat[i].M;
q.C = p.C + boat[i].C;
q.B = p.B + boat[i].B;
}
if (judge_Node(q) && !exit_open(q) && !exit_close(q)) {//避免死循环 已经扩展过的结点不再扩展
q.g = p.g + 1;
q.h = q.M + q.C - 2 * q.B;
q.f = q.g + q.h;
int pos = find(tree.begin(), tree.end(), p)-tree.begin();
q.parent = pos;
printf("扩展出新的子结点:");
printf("(%d,%d,%d,%d,%d,%d)
",q.M,q.C,q.B,q.g,q.h,q.f);
add_open(q);
tree.push_back(q);
}
else {
printf(" ------节点(%d,%d,%d)不满足条件，扩展失败------
", q.M, q.C, q.B);
}
}
printf(" =======================================================
");
add_close(p);
printf(" ******open表状态******
");
show_open();
printf(" ******close表状态******
");
show_close();
printf("------------------------------------------------------------------------------------
");
}
bool destination(Node p) {//判断p是否为目标节点
if (p.M == 0 && p.C == 0 && p.B == 0) return true;
else return false;
}
Node solve() {
Node p{ 5, 5, 1, 0, 10, 10,-1 };
open.push_back(p);
tree.push_back(p);
char c;
Node x;
while (open.size() != 0) {
x = *(open.end() - 1);//从open表中取出一个进行扩展
if (destination(x)) return x;//如果是目标状态 则结束
out_open();//从open中删除
expand(x);//扩展该结点
//getchar();
}
//return NULL;
}
void path(Node p) {
vector<Node> temp;
while (p.parent!=-1) {
temp.push_back(p);
p = tree[p.parent];
}
temp.push_back(p);
vector<Node>::iterator ite1 = temp.end() - 1;
for (; ite1 >= temp.begin(); ite1--) {
printf("(%d,%d,%d,%d,%d,%d)
", (*ite1).M, (*ite1).C, (*ite1).B, (*ite1).g, (*ite1).h, (*ite1).f);
if (ite1 == temp.begin()) break;
}
}
int main() {
Node goal=solve();
printf("求得传教士与野人过河问题状态空间的一个解为：
");
path(goal);
}