POJ3414 Pots
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题目:
给你两个容器,分别能装下A升水和B升水,并且可以进行以下操作
FILL(i) 将第i个容器从水龙头里装满(1 ≤ i ≤ 2);
DROP(i) 将第i个容器抽干
POUR(i,j) 将第i个容器里的水倒入第j个容器(这次操作结束后产生两种结果,一是第j个容器倒满并且第i个容器依旧有剩余,二是第i个容器里的水全部倒入j中,第i个容器为空)
现在要求你写一个程序,来找出能使其中任何一个容器里的水恰好有C升,找出最少操作数并给出操作过程
输入:
有且只有一行,包含3个数A,B,C(1<=A,B<=100,C<=max(A,B))
输出:
第一行包含一个数表示最小操作数K
随后K行每行给出一次具体操作,如果有多种答案符合最小操作数,输出他们中的任意一种操作过程,如果你不能使两个容器中的任意一个满足恰好C升的话,输出“impossible”
样例:
分析:简单的BFS,难点在于回溯,给每个状态用数组记录路径
#include<iostream> #include<sstream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<algorithm> #include<functional> #include<iomanip> #include<numeric> #include<cmath> #include<queue> #include<vector> #include<set> #include<cctype> #define PI acos(-1.0) const int INF = 0x3f3f3f3f; const int NINF = -INF - 1; typedef long long ll; using namespace std; int a, b, c; int used[105][105]; struct node int x, y; int flag; int path[1005];//数组中0-5分别表示6种不同操作 st; string print[6] = "FILL(1)", "FILL(2)", "DROP(1)", "DROP(2)", "POUR(1,2)", "POUR(2,1)"; void bfs() queue<node> q; for (int i = 0; i <= a; ++i) for (int j = 0; j <= b; ++j) used[i][j] = INF; memset(used, 0, sizeof(used)); st.x = 0, st.y = 0; st.flag = 0; memset(st.path, -1, sizeof(st.path)); q.push(st); used[st.x][st.y] = 1; while (q.size()) node temp = q.front(); q.pop(); if (temp.x == c || temp.y == c) cout << temp.flag << endl; for (int i = 0; i < temp.flag; ++i) cout << print[temp.path[i]] << endl; return; for (int i = 0; i < 6; ++i) node now = temp; now.flag++; if (i == 0 && now.x != a) now.x = a; if (!used[now.x][now.y]) used[now.x][now.y] = 1; now.path[temp.flag] = 0; q.push(now); else if (i == 1 && now.y != b) now.y = b; if (!used[now.x][now.y]) used[now.x][now.y] = 1; now.path[temp.flag] = 1; q.push(now); else if (i == 2 && now.x != 0) now.x = 0; if (!used[now.x][now.y]) used[now.x][now.y] = 1; now.path[temp.flag] = 2; q.push(now); else if (i == 3 && now.y != 0) now.y = 0; if (!used[now.x][now.y]) used[now.x][now.y] = 1; now.path[temp.flag] = 3; q.push(now); else if (i == 4) if (now.x + now.y > b) now.x -= b - now.y; now.y = b; else now.y += now.x; now.x = 0; if (!used[now.x][now.y]) used[now.x][now.y] = 1; now.path[temp.flag] = 4; q.push(now); else if (i == 5) if (now.x + now.y > a) now.y -= a - now.x; now.x = a; else now.x += now.y; now.y = 0; if (!used[now.x][now.y]) used[now.x][now.y] = 1; now.path[temp.flag] = 5; q.push(now); cout << "impossible" << endl; int main() cin >> a >> b >> c; bfs(); return 0;
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