10603 Fill (BFS)

Problem D Fill

There are three jugs with a volume of a, b and c liters. (a, b, and c are positive integers not greater than 200). The first and the second jug are initially empty, while the third

is completely filled with water. It is allowed to pour water from one jug into another until either the first one is empty or the second one is full. This operation can be performed zero, one or more times.

 

You are to write a program that computes the least total amount of water that needs to be poured; so that at least one of the jugs contains exactly d liters of water (d is a positive integer not greater than 200). If it is not possible to measure d liters this way your program should find a smaller amount of water d' < d which is closest to d and for which d' liters could be produced. When d' is found, your program should compute the least total amount of poured water needed to produce d' liters in at least one of the jugs.

 

Input

The first line of input contains the number of test cases. In the next T lines, T test cases follow. Each test case is given in one line of input containing four space separated integers - a, b, c and d.

 

Output

The output consists of two integers separated by a single space. The first integer equals the least total amount (the sum of all waters you pour from one jug to another) of poured water. The second integer equals d, if d liters of water could be produced by such transformations, or equals the closest smaller value d' that your program has found.

 

Sample Input

Sample Output

2

2 3 4 2

96 97 199 62

2 2

9859 62

 

Problem source: Bulgarian National Olympiad in Informatics 2003

Problem submitter: Ivaylo Riskov

Problem solution: Ivaylo Riskov, Sadrul Habib Chowdhury

题目大意：有三个杯子，容量分别为a,b,c,一开始,a,b都是空的，只有c是满的，现在开始倒水，一次倒水结束的条件是倒水的杯子倒完或者被倒的杯子倒满，问要使得至少一个杯子有恰好d单位的水，最少的倒水量，如果d不可能，那么就输出最大的小于d的数dd,和最小倒水量 

解题思路：只有三个杯子，用bfs搜索达到d的最少水量，状态得开二维，不然会超时。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#define inf 0x3f3f3f3fusing namespace std;struct jug{    int L[3],s;    friend bool operator<(jug a,jug b){        return a.s>b.s;    }};int a,b,c,d;int vis[210][210];int S[3],cap[3],TS[3];int mark;int bfs(int d){    queue<jug> q;    jug f;    f.L[0]=0,f.L[1]=0,f.L[2]=c,f.s=0;    q.push(f);    vis[0][0]=1;    int ret=inf;    while(!q.empty()){        jug x=q.front();        q.pop();        if(x.L[0]==d||x.L[1]==d||x.L[2]==d){            if(x.s<ret) ret=x.s;            continue;        }        if(x.s>=ret) continue;        for(int i=0;i<3;i++){            if(!x.L[i]) continue;            for(int j=0;j<3;j++){                if(i==j) continue;                if(x.L[j]==cap[j]) continue;                jug tmp=x;                if(x.L[i]+x.L[j]>cap[j]){                    tmp.L[i]=x.L[i]+x.L[j]-cap[j];                    tmp.L[j]=cap[j];                    tmp.s+=cap[j]-x.L[j];                }else{                    tmp.L[i]=0;                    tmp.L[j]=x.L[i]+x.L[j];                    tmp.s+=x.L[i];                }                if(vis[tmp.L[0]][tmp.L[1]]!=mark){                    vis[tmp.L[0]][tmp.L[1]]=mark;                    q.push(tmp);                }            }        }    }    mark++;    return ret;}void solve(){    cap[0]=a,cap[1]=b,cap[2]=c;    if(d>max(max(a,b),c)) {printf("0 0\n");return;}    memset(vis,-1,sizeof vis);    mark=0;    int ret=bfs(d);    if(ret!=inf) printf("%d %d\n",ret,d);    else{        for(int i=d-1;i>=0;i--){            ret=bfs(i);            if(ret!=inf){                printf("%d %d\n",ret,i);                break;            }        }    }}int main(){    int T;    scanf("%d",&T);    for(int ca=1;ca<=T;ca++){        scanf("%d%d%d%d",&a,&b,&c,&d);        solve();    }    return 0;}