# AI jobs: three wild people and three missionaries

Source: Internet
Author: User

/* <Br/> when three wild people and three priests cross the river, the number of wild people cannot exceed the number of priests at any time. <Br/> <SPAN class = 'wp _ keywordlink '> algorithm </span>: <br/> wide search for poor performance. A * does not know how to write... <Br/> <SPAN class = 'wp _ keywordlink '> Program </span> implements a solution that selects the minimum number of entries, not all feasible solutions. <Br/> */<br/> # include <iostream> <br/> # include <deque> <br/> # include <algorithm> <br/> using namespace std; <br/> struct point {<br/> int A, B, S, F; // A indicates the number of people on the departure shore, and B indicates the number of wild people, the number of people on the other side can be obtained through a simple operation. S indicates the current side of the bank, 0 indicates the departure bank, 1 indicates the opposite bank, and F indicates the location of the parent node <br/> point (int A, int B, int S, int F): A (a), B (B), S (s), F (f) {}< br/> }; <br/> deque <point> DQ; // query queue; <br/> bool operator = (const point & P1, const point & p2) {return p1.a = p2.a & p1. B = p2. B & p1.s = p2.s;} <br/> ostream & operator <(ostream & _ OS, const point & P) {// output status <br/> return _ OS <"[" <p. A <"," <p. B <"," <p.s <"," <p. F <"]"; <br/>}< br/> bool ishave (int A, int B, int s) {// judge Whether the current status has exists <br/> If (find (DQ. Begin (), DQ. End (), point (A, B, S, 0 ))! = DQ. end () return 1; <br/> return 0; <br/>}< br/> bool isgo (int A, int B, int S) // used to determine when the time can pass <br/> {<br/> If (S = 0) {<br/> if (a <0 | B <0 | A> 3 | B> 3) return 0; <br/> if (! = 0 & A <B) Return 0; <br/> return 1; <br/>}else {<br/> int a1 = 3-A, b1 = 3-B; <br/> If (A1 <0 | B1 <0 | A1> 3 | B1> 3) return 0; <br/> If (A1! = 0 & A1 <B1) return 0; <br/> return 1; <br/>}< br/> // display the path with the minimum number of moves <br/> void show (int pos) <br/> {<br/> If (DQ [POS]. F = POS) {<br/> cout <DQ [POS] <Endl; <br/> return; <br/>}< br/> show (DQ [POS]. f); <br/> cout <DQ [POS] <Endl; <br/>}< br/> int main () <br/>{< br/> int A = 3, B = 3, S; <br/> int Pos = 0; // The current node location <br/> DQ. push_back (point (a, B, 0, 0); <br/> while (Pos! = DQ. size () {// Add a loop when there are other elements that can be added <br/> A = DQ [POS]. a; <br/> B = DQ [POS]. b; <br/> S = DQ [POS]. s; <br/> if (a = 0 & B = 0 & s = 1) break; <br/> If (DQ [POS]. S = 0) {// start, listing in five cases <br/> if (a <B &! = 0) {pos ++; continue;} // status unavailable <br/> If (isgo (A, B-2, 0 )&&! Ishave (A, B-2, 1) <br/> DQ. push_back (point (A, B-2, 1, POS); <br/> If (isgo (A-2, B, 0 )&&! Ishave (A-2, B, 1) <br/> DQ. push_back (point (A-2, B, 1, POS); <br/> If (isgo (A-1, B-1, 0 )&&! Ishave (A-1, B-1, 1) <br/> DQ. push_back (point (A-1, B-1, 1, POS); <br/> If (isgo (A-1, B, 0 )&&! Ishave (A-1, B, 1) <br/> DQ. push_back (point (A-1, B, 1, POS); <br/> If (isgo (A, B-1, 0 )&&! Ishave (A, B-1, 1) <br/> DQ. push_back (point (A, B-1, 1, POS); <br/>}else {// back <br/> If (3-)! = 0 & (3-a) <(3-B) {pos ++; continue;} // The status is unavailable <br/> If (isgo (a + 2, B, 1 )&&! Ishave (a + 2, B, 0) <br/> DQ. push_back (point (a + 2, B, 0, POS); <br/> If (isgo (A, B + 2, 1 )&&! Ishave (A, B + 2, 0) <br/> DQ. push_back (point (A, B + 2, 0, POS); <br/> If (isgo (a + 1, B + 1, 1, 1 )&&! Ishave (a + 1, B + 1, 0) <br/> DQ. push_back (point (a + 1, B + 1, 0, POS); <br/> If (isgo (a + 1, B, 1 )&&! Ishave (a + 1, B, 0) <br/> DQ. push_back (point (a + 1, B, 0, POS); <br/> If (isgo (A, B + 1, 1, 1 )&&! Ishave (A, B + 1, 0) <br/> DQ. push_back (point (A, B + 1, 0, POS); <br/>}< br/> POS ++; <br/>}< br/>/* queue status */<br/> for (INT I = 0; I <DQ. size (); I ++) <br/> cout <I <":" <DQ [I] <Endl; <br/> // display an optimal path <br/> cout <"--------------------" <Endl; <br/> show (POS); <br/>}

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