B - Fill-白红宇

B - Fill

阅读量：6659 次

发布时间：2019-06-25

本文共 3348 字，大约阅读时间需要 11 分钟。

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

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Appoint description:

Description

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 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

倒水问题状态bfs搜索

1 #include 
       
         2 #include 
        
          3 #include 
         
           4 #include 
          
            5 #include 
           
             6 #include 
            
              7 #include 
             
               8 #include 
              
                9 #include 
               10 #include 
                
                 11 #include 
                 
                  12 #include 
                  
                   13 #include 
                   
                    14 using namespace std;15 const int INF=0x5fffffff;16 const int EXP=1e-5;17 const int MS=205;18 struct node19 {20 int v[3],dist; //这里的dist表示到达这个状态所倒的水量21 bool operator < (const node &a)const22 {23 return dist>a.dist;24 }25 };26 27 int vis[MS][MS],cap[3],ans[MS];28 29 void updata(const node &u)30 {31 for(int i=0;i<3;i++)32 {33 int d=u.v[i]; // 对每个体积进行更新34 if(ans[d]<0||u.dist
                    
                      que;45 node start;46 start.dist=0;47 start.v[0]=0;48 start.v[1]=0;49 start.v[2]=c;50 que.push(start);51 while(!que.empty())52 {53 node u=que.top();54 que.pop();55 updata(u);56 if(ans[d]>=0)57 break;58 for(int i=0;i<3;i++)59 for(int j=0;j<3;j++) //i-->j;60 if(i!=j)61 {62 if(u.v[i]==0||u.v[j]==cap[j])63 continue; //i没有水 or j已经满了64 int cnt=min(cap[j],u.v[i]+u.v[j])-u.v[j];65 node t;66 memcpy(&t,&u,sizeof(u)); // t=u;67 t.dist=u.dist+cnt;68 t.v[i]-=cnt;69 t.v[j]+=cnt;70 if(!vis[t.v[0]][t.v[1]])71 {72 vis[t.v[0]][t.v[1]]=1;73 que.push(t);74 }75 }76 }77 while(d>=0)78 {79 if(ans[d]>=0)80 {81 printf("%d %d\n",ans[d],d);82 return ;83 }84 d--;85 }86 }87 88 int main()89 {90 int T,a,b,c,d;91 scanf("%d",&T);92 while(T--)93 {94 scanf("%d%d%d%d",&a,&b,&c,&d);95 solve(a,b,c,d);96 }97 return 0;98 }

转载于:https://www.cnblogs.com/767355675hutaishi/p/4330524.html

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