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HDU 1495 非常可乐 (BFS)
题意:略。
析:由于只有三只杯子,那么我们可以用两个杯子的状态,那么第三只的状态也可以确定下来,每次倒水,要么全倒过去,要么把那个杯子倒满。
注意题意是要保证最后是两个一样多。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 200 + 10; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int d[maxn][maxn]; int all; struct Node{ int a, b; Node(int aa, int bb) : a(aa), b(bb) { } }; bool judge(Node &u){ if(u.a * 2 == all) return u.b == 0 || (all - u.a - u.b) == 0; if(u.b * 2 == all) return u.a == 0 || (all - u.a - u.b) == 0; if((all - u.a - u.b) * 2 == all) return u.a == 0 || u.b == 0; return false; } void update(int &x, int &y, Node &u, queue<Node> &q){ d[x][y] = d[u.a][u.b] + 1; q.push(Node(x, y)); } int bfs(){ queue<Node> q; q.push(Node(all, 0)); memset(d, -1, sizeof d); d[all][0] = 0; while(!q.empty()){ Node u = q.front(); q.pop(); if(judge(u)) return d[u.a][u.b]; // pour the first to the second or not int t = min(u.a, n - u.b); int x = u.a - t, y = u.b + t; if(d[x][y] == -1) update(x, y, u, q); t = min(all-u.a, u.b); x = u.a + t, y = u.b - t; if(d[x][y] == -1) update(x, y, u, q); // pour the first to the third or not t = min(u.a, m - (all-u.a-u.b)); x = u.a - t, y = u.b; if(d[x][y] == -1) update(x, y, u, q); t = min(all-u.a, all-u.a-u.b); x = u.a + t; if(d[x][y] == -1) update(x, y, u, q); //pour the second to the third or not t = min(u.b, m - (all-u.a-u.b)); x = u.a, y = u.b - t; if(d[x][y] == -1) update(x, y, u, q); t = min(n-u.b, all-u.a-u.b); y = u.b + t; if(d[x][y] == -1) update(x, y, u, q); } return -1; } int main(){ while(scanf("%d %d %d", &all, &n, &m) == 3 && all + n + m){ all <<= 1, n <<= 1, m <<= 1; int ans = bfs(); if(ans == -1) printf("NO\n"); else printf("%d\n", ans); } return 0; }
HDU 1495 非常可乐 (BFS)
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