题目链接
http://poj.org/problem?id=3414
题意 给出两个杯子 容量分别为 A B 然后给出C 是目标容量 有三种操作 1 将一个杯子装满 2.将一个杯子全都倒掉 3.将一个杯子的水倒到另一个杯子里面
如果某个杯子里面的水 能够达到 目标容量 那么就输出步骤
思路 BFS 并且要存储步骤
每一步一共有六步操作 记得标记
AC代码
#include <cstdio> #include <cstring> #include <ctype.h> #include <cstdlib> #include <cmath> #include <climits> #include <ctime> #include <iostream> #include <algorithm> #include <deque> #include <vector> #include <queue> #include <string> #include <map> #include <stack> #include <set> #include <numeric> #include <sstream> #include <iomanip> #include <limits> #define CLR(a) memset(a, 0, sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; typedef pair <int, int> pii; typedef pair <ll, ll> pll; typedef pair<string, int> psi; typedef pair<string, string> pss; const double PI = acos(-1.0); const double E = exp(1.0); const double eps = 1e-8; const int INF = 0x3f3f3f3f; const int maxn = 1e2 + 5; const int MOD = 1e9 + 7; /* 0 FILL 1 1 FILL 2 2 DROP 1 3 DROP 2 4 POUR 1 2 5 POUR 2 1 */ int visit[maxn][maxn]; vector <int> ans; int flag; int a, b, c; struct node { int a, b; vector <int> v; }; void bfs() { queue <node> q; node tmp; tmp.a = 0; tmp.b = 0; tmp.v.clear(); q.push(tmp); visit[0][0] = 1; while (!q.empty()) { node u = q.front(), v; q.pop(); if (u.a == c || u.b == c) { flag = 1; ans = u.v; return; } if (u.a < a) { v.a = a; v.b = u.b; if (visit[v.a][v.b] == 0) { v.v = u.v; v.v.pb(0); q.push(v); visit[v.a][v.b] = 1; } } if (u.b < b) { v.a = u.a; v.b = b; if (visit[v.a][v.b] == 0) { v.v = u.v; v.v.pb(1); q.push(v); visit[v.a][v.b] = 1; } } if (u.a > 0) { v.a = 0; v.b = u.b; if (visit[v.a][v.b] == 0) { v.v = u.v; v.v.pb(2); q.push(v); visit[v.a][v.b] = 1; } } if (u.b > 0) { v.a = u.a; v.b = 0; if (visit[v.a][v.b] == 0) { v.v = u.v; v.v.pb(3); q.push(v); visit[v.a][v.b] = 1; } } if (u.a < a) { int c = a - u.a; if (u.b >= c) { v.b = u.b - c; v.a = a; } else { v.b = 0; v.a = u.a + u.b; } if (visit[v.a][v.b] == 0) { v.v = u.v; v.v.pb(5); q.push(v); visit[v.a][v.b] = 1; } } if (u.b < b) { int c = b - u.b; if (u.a >= c) { v.a = u.a - c; v.b = b; } else { v.a = 0; v.b = u.a + u.b; } if (visit[v.a][v.b] == 0) { v.v = u.v; v.v.pb(4); q.push(v); visit[v.a][v.b] = 1; } } } } int main() { map <int, string> M; M[0] = "FILL(1)"; M[1] = "FILL(2)"; M[2] = "DROP(1)"; M[3] = "DROP(2)"; M[4] = "POUR(1,2)"; M[5] = "POUR(2,1)"; CLR(visit, 0); scanf("%d%d%d", &a, &b, &c); flag = 0; bfs(); if (flag == 0) printf("impossible\n"); else { int len = ans.size(); cout << len << endl; for (int i = 0; i < len; i++) cout << M[ans[i]] << endl; } }转载于:https://www.cnblogs.com/Dup4/p/9433093.html
