poj 3414 pots [经典 BFS ]

最新推荐文章于 2020-09-10 11:11:10 发布

原创最新推荐文章于 2020-09-10 11:11:10 发布 · 447 阅读

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本文探讨了智能算法在大数据开发领域的应用，包括Hadoop、Spark等技术的使用及优化策略。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <memory.h>
#include <cstdio>
#define Bug cout << "here\n";
using namespace std;
const int N = 101;

struct node {
    int k1, k2, steps;
    int op;
    node *pre;
}que[N*N];
node cur, nex;
int vis[N][N];

char str[][10] = {"FILL(1)", "DROP(1)", "FILL(2)", "DROP(2)", "POUR(1,2)", "POUR(2,1)"};

void DFS(node* t) {
    if(t->pre != NULL) {
        DFS(t->pre);
        cout << str[t->op] << endl;
    }
}
void BFS(int v1, int v2, int c) {
    que[0] = cur;
    vis[0][0] = 1; int yi;
    int head = 0, tail = 1;
    node* cp;
    while(head < tail) {
        cp = &que[head++];
        if(cp->k1 == c || cp->k2 == c) {
            printf("%d\n", cp->steps);
            DFS(cp);
            return;
        }
        for(int i = 0; i < 6; i++) {
            switch(i) {
                case 0 : nex.k1 = v1; nex.k2 = cp->k2; break;  // A 满
                case 1 : nex.k1 = 0; nex.k2 = cp->k2; break;   // A 空
                case 2 : nex.k1 = cp->k1; nex.k2 = v2; break;  // B 满
                case 3 : nex.k1 = cp->k1; nex.k2 = 0; break;   // B 空
                case 4 : yi = cp->k1 + cp->k2 - v2;
                         if(yi > 0) {
                             nex.k1 = yi;
                             nex.k2 = v2;
                         }
                         else {
                              nex.k1 = 0;
                              nex.k2 = yi+v2;
                         }
                         break;
                case 5 : yi = cp->k1 + cp->k2 - v1;
                         if(yi > 0) {
                            nex.k1 = v1;
                            nex.k2 = yi;
                         }
                         else {
                            nex.k1 = yi+v1;
                            nex.k2 = 0;
                         }
            }
            nex.steps = cp->steps + 1;
            nex.pre = cp;
            nex.op = i;
            if(!vis[nex.k1][nex.k2]) {
                que[tail++] = nex;
                vis[nex.k1][nex.k2] = 1;
            }
        }
    }
    cout << "impossible" << endl;
}

int main() {
    int v1, v2, c;
    while(scanf("%d%d%d", &v1, &v2, &c) == 3) {
        memset(vis, 0, sizeof(vis));
        BFS(v1, v2, c);
    }
    return 0;
}