POJ3414 Pots

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
using namespace std;

struct node{
    int a,b,t;
    bool operator < (const node &tmp)const{
        return t > tmp.t;
    }
};
struct path{
    int x,y;
}pa[105][105];
int a,b,c;
int vis[105][105];
priority_queue<node> q;

void solve(int a,int b,node cur,int mode){
    if(!vis[a][b]){
        vis[a][b] = mode;
        pa[a][b].x = cur.a;
        pa[a][b].y = cur.b;
        q.push(node{a,b,cur.t+1});
    }
}

node bfs(){
    node cur;
    while(!q.empty())
        q.pop();
    memset(vis,0,sizeof(vis));
    vis[0][0] = -1;
    pa[0][0].x = -1;
    pa[0][0].y = -1;
    q.push(node{0,0,0});
    while(!q.empty()){
        cur = q.top();
        q.pop();
        if(cur.a == c || cur.b == c){
            return cur;
        }
        //一共６种状态
        //1.fill:a
        solve(a,cur.b,cur,1);
        //2.fill:b
        solve(cur.a,b,cur,2);
        //3.drop:a
        solve(0,cur.b,cur,3);
        //4.drop:b
        solve(cur.a,0,cur,4);
        //5.pour:a->b
        if(cur.a >= b-cur.b){
            solve(cur.a-(b-cur.b),b,cur,5);
        }
        else{
            solve(0,cur.a+cur.b,cur,5);
        }
        //6.pour:b->a
        if(cur.b >= a-cur.a){
            solve(a,cur.b-(a-cur.a),cur,6);
        }
        else{
            solve(cur.a+cur.b,0,cur,6);
        }

    }
    return node{-1,-1,-1};
}

void printpath(int x,int y,int t){
    if(!t)
        return ;
    printpath(pa[x][y].x,pa[x][y].y,t-1);
    switch(vis[x][y]){
        case 1:
            printf("FILL(1)\n");
			break;
		case 2:
			printf("FILL(2)\n");
			break;
		case 3:
			printf("DROP(1)\n");
			break;
		case 4:
			printf("DROP(2)\n");
			break;
		case 5:
			printf("POUR(1,2)\n");
			break;
		case 6:
			printf("POUR(2,1)\n");
			break;
    }
}

int main(){
    while(~scanf("%d%d%d",&a,&b,&c)){
        node ans = bfs();
        if(ans.t < 0)
            printf("impossible\n");
        else{
            printf("%d\n",ans.t);
            printpath(ans.a,ans.b,ans.t);
        }
    }
    return 0;
}