[kuangbin带你飞]简单搜索:Pots

[POJ-3414]
You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
DROP(i) empty the pot i to the drain;
POUR(i,j) pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).
Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cstdlib>
#include<cmath>
using namespace std;
#define ll long long
#define lb long double
#define INF 0x3f3f3f3f
#define maxn 101 * 101 + 5
bool book[maxn];
int n, m, k, ans = -1, ex, ey, cnt, ei, ej;
struct node{
    int x;
    int y;
    int d;
    int vi;
    int vj;
}last[105][105], anss[105];
int f(int a, int b){
    return a * 101 + b;
}
void bfs(){
    memset(book, 0, sizeof(book));
    book[0] = 1;
    node st;
    st.x = 0; st.y = 0; st.d = 0;
    queue<node> que;
    que.push(st);
    while(!que.empty()){
        node cur = que.front();
        que.pop();
        if(cur.x == k || cur.y == k){
            ans = cur.d; ex = cur.x; ey = cur.y; ei = cur.vi; ej = cur.vj;
            break;
        }
        for(int i = 1 ; i <= 3 ; ++ i){
            node temp;
            if(i == 1){
                for(int j = 1 ; j <= 2 ; ++ j){
                    if(j == 1){//FILL(1)
                        temp.x = n; temp.y = cur.y;
                    }
                    if(j == 2){//FILL(2)
                        temp.x = cur.x; temp.y = m;
                    }
                    if(!book[f(temp.x, temp.y)]){
                        temp.d = cur.d + 1; book[f(temp.x, temp.y)] = 1; temp.vi = i; temp.vj = j; que.push(temp);
                        last[temp.x][temp.y].x = cur.x; last[temp.x][temp.y].y = cur.y;
                        last[temp.x][temp.y].vi = i; last[temp.x][temp.y].vj = j;
                    }
                }
            }
            if(i == 2){//POUR(2,1)
                for(int j = 1 ; j <= 2 ; ++ j){
                    if(j == 1){//POUR(1,2)
                        if(cur.x <= m - cur.y){
                            temp.x = 0; temp.y = cur.x + cur.y;
                        }
                        else{
                            temp.x = cur.x - (m - cur.y); temp.y = m;
                        }
                    }
                    if(j == 2){//POUR(2,1)
                        if(cur.y <= n - cur.x){
                            temp.y = 0; temp.x = cur.y + cur.x;
                        }
                        else{
                            temp.y = cur.y - (n - cur.x); temp.x = n;
                        }
                    }
                    if(!book[f(temp.x, temp.y)]){
                        temp.d = cur.d + 1; book[f(temp.x, temp.y)] = 1; temp.vi = i; temp.vj = j; que.push(temp);
                        last[temp.x][temp.y].x = cur.x; last[temp.x][temp.y].y = cur.y;
                        last[temp.x][temp.y].vi = i; last[temp.x][temp.y].vj = j;
                    }
                }
            }
            if(i == 3){//DROP(1)
                for(int j = 1 ; j <= 2 ; ++ j){
                    if(j == 1){
                        temp.x = 0; temp.y = cur.y;
                    }
                    if(j == 2){
                        temp.x = cur.x; temp.y = 0;
                    }
                    if(!book[f(temp.x, temp.y)]){
                        temp.d = cur.d + 1; book[f(temp.x, temp.y)] = 1; temp.vi = i; temp.vj = j; que.push(temp);
                        last[temp.x][temp.y].x = cur.x; last[temp.x][temp.y].y = cur.y;
                        last[temp.x][temp.y].vi = i; last[temp.x][temp.y].vj = j;
                    }
                }
            }

            //记录路径 i, j
            //last[temp.x][temp.y].x = i; last[temp.x][temp.y].y = j;
        }
    }
}
int main()
{
    scanf("%d %d %d",  &n, &m, &k);
    bfs();

    //anss[++cnt].x = ex; anss[cnt].y = ey; anss[cnt].vi = ei; anss[cnt].vj = ej;
    while(ex !=0 || ey != 0){
        int tempx = ex, tempy = ey;
        ex = last[tempx][tempy].x; ey = last[tempx][tempy].y; ei = last[tempx][tempy].vi; ej = last[tempx][tempy].vj;
        anss[++ cnt].x = ex; anss[cnt].y = ey; anss[cnt].vi = ei; anss[cnt].vj = ej;
    }
    if(ans == -1) cout << "impossible" << endl;
    else{
        cout << ans << endl;
        for(int i = cnt ; i >= 1 ; i --){
            if(anss[i].vi == 1){
                if(anss[i].vj == 1) cout << "FILL(1)" << endl;
                if(anss[i].vj == 2) cout << "FILL(2)" << endl;
            }
            if(anss[i].vi == 2){
                if(anss[i].vj == 1) cout << "POUR(1,2)" << endl;
                if(anss[i].vj == 2) cout << "POUR(2,1)" << endl;
            }
            if(anss[i].vi == 3){
                if(anss[i].vj == 1) cout << "DROP(1)" << endl;
                if(anss[i].vj == 2) cout << "DROP(2)" << endl;
            }
        }
    }

    return 0;
}