kuangbin 简单搜索 H题

H - Pots

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.

Input
On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).

Output
The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.

Sample Input
3 5 4
Sample Output
6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)

题解：

bfs找最短路,dfs回溯求出路径。
代码量有点长，下次优化一下，不过0ms可过。

代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>using namespace std;const int MAXN = 110;const int INF = 0x3f3f3f3f;struct Status{    int x,y;    Status(int x,int y):x(x),y(y){}    Status(){}};int A,B,C;char op[6][10]={{"FILL(1)"},{"FILL(2)"},{"DROP(1)"},{"DROP(2)"},{"POUR(1,2)"},{"POUR(2,1)"}};typedef struct{    int op,steps;}Vis;Vis vis[MAXN][MAXN];vector<string> ans;bool flag=0;void dfs(int x,int y,int times){     if(flag) return;     if(times==0&&(x==C||y==C))     {        for(int i=0;i<ans.size();i++)        {            cout<<ans[i]<<endl;        }        flag=1;        return;     }     //FILL(1)     int nx = A,ny = y;     if(vis[nx][ny].steps==vis[x][y].steps+1)     {         ans.push_back(op[vis[nx][ny].op]);         dfs(nx,ny,times-1);         ans.pop_back();     }     //FILL(2)      nx = x,ny = B;     if(vis[nx][ny].steps==vis[x][y].steps+1)     {         ans.push_back(op[vis[nx][ny].op]);         dfs(nx,ny,times-1);         ans.pop_back();     }     //DROP(1)      nx=0,ny=y;     if(vis[nx][ny].steps==vis[x][y].steps+1)     {         ans.push_back(op[vis[nx][ny].op]);         dfs(nx,ny,times-1);         ans.pop_back();     }     //DROP(2)      nx=x,ny=0;     if(vis[nx][ny].steps==vis[x][y].steps+1)     {         ans.push_back(op[vis[nx][ny].op]);         dfs(nx,ny,times-1);         ans.pop_back();     }     //POUR(1,2)        if(x<=B-y)        {           nx=0;           ny=y+x;        }        else        {            nx=x-(B-y);            ny=B;        }     if(vis[nx][ny].steps==vis[x][y].steps+1)     {         ans.push_back(op[vis[nx][ny].op]);         dfs(nx,ny,times-1);         ans.pop_back();     }     //POUR(2,1)       if(y<=A-x)        {           nx=y+x;           ny=0;        }        else        {            nx=A;            ny=y-(A-x);        }      if(vis[nx][ny].steps==vis[x][y].steps+1)     {         ans.push_back(op[vis[nx][ny].op]);         dfs(nx,ny,times-1);         ans.pop_back();     }}void bfs(){    memset(vis,0x3f,sizeof(vis));    queue<Status> que;    que.push(Status(0,0));    vis[0][0].steps=0;    while(que.size())    {        Status start = que.front();        que.pop();        if(start.x==C||start.y==C)        {            printf("%d\n",vis[start.x][start.y].steps);            flag=0;            dfs(0,0,vis[start.x][start.y].steps);            return;        }        Status temp;        //FILL(1)        temp.x=A,temp.y=start.y;        if(vis[A][temp.y].steps==INF)        {            vis[A][temp.y].steps=vis[start.x][start.y].steps+1;            vis[A][temp.y].op=0;            que.push(temp);        }        //FILL(2)         temp.x=start.x,temp.y=B;        if(vis[temp.x][B].steps==INF)        {            vis[temp.x][B].steps=vis[start.x][start.y].steps+1;            vis[temp.x][B].op=1;            que.push(temp);        }        //DROP(1)        temp.x=0,temp.y=start.y;        if(vis[0][temp.y].steps==INF)        {           vis[0][temp.y].steps=vis[start.x][start.y].steps+1;           vis[0][temp.y].op=2;           que.push(temp);        }        //DROP(2)        temp.x=start.x,temp.y=0;        if(vis[temp.x][0].steps==INF)        {            vis[temp.x][0].steps=vis[start.x][start.y].steps+1;            vis[temp.x][0].op=3;            que.push(temp);        }        //POUR(1,2)        if(start.x<=B-start.y)        {           temp.x=0;           temp.y=start.y+start.x;        }        else        {            temp.x=start.x-(B-start.y);            temp.y=B;        }        if(vis[temp.x][temp.y].steps==INF)        {           vis[temp.x][temp.y].steps=vis[start.x][start.y].steps+1;            vis[temp.x][temp.y].op=4;            que.push(temp);        }        //POUR(2,1)         if(start.y<=A-start.x)        {           temp.x=start.y+start.x;           temp.y=0;        }        else        {            temp.x=A;            temp.y=start.y-(A-start.x);        }        if(vis[temp.x][temp.y].steps==INF)        {           vis[temp.x][temp.y].steps=vis[start.x][start.y].steps+1;            vis[temp.x][temp.y].op=5;            que.push(temp);        }    }    printf("impossible\n");}int main(){    while(cin>>A>>B>>C)    {        bfs();        ans.clear();    }    return 0;}