Poj 3414 Pots (bfs)

Link: poj 3414

Question: The capacity of two bottles A, B, and A target water volume C are given. You can perform the following operations on A and B:

FILL (I) fills the bottle with water

DROP (I) empty the bottle I

POUR (I, j) POUR the water in bottle I into bottle j. After this operation, either the bottle j is filled with water or the bottle I is empty

Perform at least several operations to make the water in A or B package C and output the specific operation.

Analysis: bfs can be used in six aspects. To output specific operations, you can use arrays to simulate queues,

Record the status of the previous operation, and then trace the path.

# Include <stdio. h> # include <string. h> # include <stdlib. h> struct stu {int step, a, B, pos;} que [1010]; int a, B, c, ope [1010], step [1010], x, y, front, rear; bool vis [105] [105]; void fill (int I) {if (I = 1) x = a; else y = B ;} void drop (int I) {if (I = 1) x = 0; else y = 0;} void pour (int I, int j) {if (I = 1) {if (x + y> = B) {x-= B-y; y = B;} else {y + = x; x = 0 ;}} else {if (x + y> = a) {y-= a-x; x = a;} else {x + = y; y = 0 ;}} void is_push (int step, I Nt num) {if (! Vis [x] [y]) {que [rear]. a = x; que [rear]. B = y; que [rear]. step = step + 1; // record the number of steps que [rear]. pos = front; // record the previous state ope [rear ++] = num; // The current operation vis [x] [y] = true; // Mark as Accessed} int bfs () {int I, num, q [7] = {0,110,120,210,220,312,321}; // The q array is used to mark the executed operation, in order to finally output struct stu t; front = 0; rear = 1; memset (vis, 0, sizeof (vis); vis [0] [0] = 1; que [front]. a = que [front]. B = 0; que [front]. step = 0; while (front <rear) {t = que [front]; if (t. a = c | t. B = c ) {Num = t. step; step [num] = front; num --; while (num) {step [num] = que [step [num + 1]. pos; num --;} return t. step;} for (I = 1; I <= 6; I ++) {x = t. a; y = t. b; switch (I) {case 1: fill (1); break; case 2: fill (2); break; case 3: drop (1); break; case 4: drop (2); break; case 5: pour (1, 2); break; case 6: pour (2, 1);} is_push (t. step, q [I]);} front ++;} return-1;} int main () {int sum, I, k; while (scanf ("% d", & a, & B, & c )! = EOF) {sum = bfs (); if (sum =-1) {printf ("impossible \ n"); continue;} printf ("% d \ n ", sum); for (I = 1; I <= sum; I ++) {k = ope [step [I]; if (k/100 = 1) printf ("FILL (% d) \ n", k/10% 10); else if (k/100 = 2) printf ("DROP (% d) \ n ", k/10% 10); else if (k/100 = 3) printf ("POUR (% d, % d) \ n", k/10% 10, k % 10) ;}} return 0 ;}

Poj 3414 Pots (bfs)