NYOJ21 三个水杯

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题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=21

题目思路 : 广搜

题目ac代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <iterator>#include <string>#include <stack>#include <queue>using namespace std;const int M = 100;bool vis[M][M][M];        //记录是否出现过此状态struct Cup{int v[3];             //用数组记录三个杯子中的状态很方便编码int step;             // 最小步数}s,e;       // 杯子大小，最终状态bool del(Cup &now, int start, int end )     // 传的是引用，start杯子向end杯子倒水{int sum = now.v[start] + now.v[end] ;if(now.v[start] != 0 && now.v[end] != s.v[end])   // 能倒水{now.v[end] = (now.v[start] >= (s.v[end] - now.v[end])) ? s.v[end] : now.v[end] + now.v[start];now.v[start] = sum - now.v[end] ;if(vis[now.v[0]][now.v[1]][now.v[2]] == false)   //没有出现过这个状态就返回true入队列{vis[now.v[0]][now.v[1]][now.v[2]] = true;return true;}}return false;}int bfs(){queue<Cup>q;struct Cup tmp = {s.v[0], 0, 0};tmp.step = 0;q.push(tmp);vis[s.v[0]][0][0] = true;while(!q.empty()){Cup tmp = q.front(),now;q.pop();now = tmp;if(now.v[0] == e.v[0] && now.v[1] == e.v[1] && now.v[2] == e.v[2])return now.step ;for(int i = 0; i < 3; i++)        //遍历倒水的六个状态{for(int j = 0; j < 3; j++){now = tmp;if( i != j && del(now,i,j)){now.step = tmp.step + 1;q.push(now);}}}}return -1; ;}int main(){int T;cin >> T;while(T --){memset(vis, false, sizeof(vis));cin >> s.v[0] >> s.v[1] >> s.v[2];cin >> e.v[0] >> e.v[1] >> e.v[2];cout << bfs() << endl;}return 0;}