USCAO section Mother's Milk（搜索）

关注

原创最新推荐文章于 2026-05-18 09:42:25 发布 847 阅读 AI 写同款· GEO 优化 ›

USACO section 3.4 Closed Fences(计算几何+叉积+二分) Closed Fences A closed fence in the plane is a set of non-crossing, connected line segments with N corners (3 i, yi}, i in (1..N). Every pair of adjacent vertices defines a side of the fence. 阅读详情

Mother's Milk

Farmer John has three milking buckets of capacity A, B, and C liters. Each of the numbers A, B, and C is an integer from 1 through 20, inclusive. Initially, buckets A and B are empty while bucket C is full of milk. Sometimes, FJ pours milk from one bucket to another until the second bucket is filled or the first bucket is empty. Once begun, a pour must be completed, of course. Being thrifty, no milk may be tossed out.

Write a program to help FJ determine what amounts of milk he can leave in bucket C when he begins with three buckets as above, pours milk among the buckets for a while, and then notes that bucket A is empty.

PROGRAM NAME: milk3

INPUT FORMAT

A single line with the three integers A, B, and C.

SAMPLE INPUT (file milk3.in)

8 9 10

OUTPUT FORMAT

A single line with a sorted list of all the possible amounts of milk that can be in bucket C when bucket A is empty.

SAMPLE OUTPUT (file milk3.out)

1 2 8 9 10

SAMPLE INPUT (file milk3.in)

2 5 10

SAMPLE OUTPUT (file milk3.out)

5 6 7 8 9 10

个人深搜解：这样编程需要很认真，否则容易出错，我也是调了挺久的

/*
ID:nealgav1
PROG:milk3
LANG:C++
*/
#include<fstream>
#include<cstring>
using namespace std;
int a,b,c;
bool s[22][22][22];
bool flag[23];
void dfs(int aa,int bb,int cc)
{
  if(s[aa][bb][cc])return;
  s[aa][bb][cc]=1;
  int x;
  if(aa==0)
  {
    flag[cc]=1;
    if(aa+cc<=a){x=cc;aa+=x;cc-=x;dfs(aa,bb,cc);aa-=x;cc+=x;}//c->a
    else {x=a-aa;cc-=x;aa+=x;dfs(aa,bb,cc);cc+=x;aa-=x;}
    if(aa+bb<=a){x=bb;bb-=x;aa+=x;dfs(aa,bb,cc);bb+=x;aa-=x;}//b->a
    else {x=a-aa;bb-=x;aa+=x;dfs(aa,bb,cc);bb+=x;aa-=x;}
    if(bb+cc<=c){x=bb;bb-=x;cc+=x;dfs(aa,bb,cc);cc-=x;bb+=x;}//b->c
    else{x=(c-cc);cc+=x;bb-=x;dfs(aa,bb,cc);bb+=x;cc-=x;}
    if(bb+cc<=b){x=cc;cc-=x;bb+=x;dfs(aa,bb,cc);cc+=x;bb-=x;}//c->b
    else{x=(b-bb);bb+=x;cc-=x;dfs(aa,bb,cc);bb-=x;cc+=x;}
  }
  else
  {
    if(aa+bb<=b){x=aa;bb+=x;aa-=x;dfs(aa,bb,cc);bb-=x;aa+=x;}//a->b
    else{x=(b-bb);aa-=x;bb+=x;dfs(aa,bb,cc);bb-=x;aa+=x;}
    if(aa+cc<=c){x=aa;cc+=x;aa-=x;dfs(aa,bb,cc);cc-=x;aa+=x;}//a->c
    else{x=(c-cc);aa-=x;cc+=x;dfs(aa,bb,cc);cc-=x;aa+=x;}
    if(aa+cc<=a){x=cc;aa+=x;cc-=x;dfs(aa,bb,cc);aa-=x;cc+=x;}//c->a
    else {x=a-aa;cc-=x;aa+=x;dfs(aa,bb,cc);cc+=x;aa-=x;}
    if(aa+bb<=a){x=bb;bb-=x;aa+=x;dfs(aa,bb,cc);bb+=x;aa-=x;}//b->a
    else {x=a-aa;bb-=x;aa+=x;dfs(aa,bb,cc);bb+=x;aa-=x;}
    if(bb+cc<=c){x=bb;bb-=x;cc+=x;dfs(aa,bb,cc);cc-=x;bb+=x;}//b->c
    else{x=(c-cc);cc+=x;bb-=x;dfs(aa,bb,cc);bb+=x;cc-=x;}
    if(bb+cc<=b){x=cc;cc-=x;bb+=x;dfs(aa,bb,cc);cc+=x;bb-=x;}//c->b
    else{x=(b-bb);bb+=x;cc-=x;dfs(aa,bb,cc);bb-=x;cc+=x;}
  }
}
int main()
{  int aa,bb,cc;
ifstream fin("milk3.in");
ofstream fout("milk3.out");
  while(fin>>a>>b>>c)
  {
    memset(flag,0,sizeof(flag));
    memset(s,0,sizeof(s));
        dfs(0,0,c);
        int ans=0;
        for(int i=0;i<21;i++)
        if(flag[i])
        ans=i;
        for(int i=0;i<21;i++)
        if(flag[i]&&ans!=i)
        fout<<i<<" ";
        fout<<ans<<"\n";
  }
}

以下摘自独孤的博客，感觉着个方法代码敲的少，不易错

还是个搜索题，每次最多无非有6种倒法，即a->b;a->c;b->a;b->c;c->a;c->b。所以如果用a、b、c代表三个桶的容积，x、y、z代表当前三个桶内的牛奶数，那么不难算出执行完a->b后，x变为max(0,x+y-b)，y变为min(b,x+y)，z不变。其它倒法类似。

最后注意判重最好的方法就是用bool数组flag[i][j][k]表示a、b、c桶分别装i、j、k升牛奶这种状态是否曾经搜到过。但由于牛奶总数不变，故知道了i和j，其实k也就知道了，所以用二维bool数组flag[i][j]就够了。

源码如下：

#include <fstream>

using namespace std;

int a,b,c;
bool flag[21][21],can[21];

void dfs(int,int,int);

int main() {
ifstream fin ("milk3.in");
ofstream fout ("milk3.out");
fin >> a >> b >> c;
dfs(0,0,c);
for (int i=0;i<c;i++) if (can[i]) fout << i << ' ';
fout << c << endl;
fin.close();
fout.close();
return 0;
}

void dfs(int x,int y,int z) {
if (flag[x][y]) return;
flag[x][y]=true;
if (x==0) can[z]=true;
if (x>0 && y<b) dfs(max(0,x+y-b),min(b,x+y),z);
if (x>0 && z<c) dfs(max(0,x+z-c),y,min(c,x+z));
if (y>0 && x<a) dfs(min(a,y+x),max(0,y+x-a),z);
if (y>0 && z<c) dfs(x,max(0,y+z-c),min(c,y+z));
if (z>0 && x<a) dfs(min(a,z+x),y,max(0,z+x-a));
if (z>0 && y<b) dfs(x,min(b,z+y),max(0,z+y-b));
}

USER: Neal Gavin Gavin [nealgav1]
TASK: milk3
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.011 secs, 3348 KB]
   Test 2: TEST OK [0.000 secs, 3348 KB]
   Test 3: TEST OK [0.000 secs, 3348 KB]
   Test 4: TEST OK [0.000 secs, 3348 KB]
   Test 5: TEST OK [0.000 secs, 3348 KB]
   Test 6: TEST OK [0.000 secs, 3348 KB]
   Test 7: TEST OK [0.000 secs, 3348 KB]
   Test 8: TEST OK [0.000 secs, 3348 KB]
   Test 9: TEST OK [0.000 secs, 3348 KB]
   Test 10: TEST OK [0.000 secs, 3348 KB]

All tests OK.
Your program ('milk3') produced all correct answers!  This is your
submission #2 for this problem.  Congratulations!

Here are the test data inputs:

------- test 1 ----
2 5 10
------- test 2 ----
20 20 20
------- test 3 ----
5 11 15
------- test 4 ----
2 12 20
------- test 5 ----
19 4 11
------- test 6 ----
5 11 13
------- test 7 ----
3 20 20
------- test 8 ----
7 16 20
------- test 9 ----
20 10 9
------- test 10 ----
7 12 18

Keep up the good work!

Mother's Milk
Russ Cox

We use a simple depth-first search to find all the possible states for the three buckets, pruning the search by not researching from states we've seen before.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>

#define MAX 20

typedef struct State	State;
struct State {
    int a[3];
};

int seen[MAX+1][MAX+1][MAX+1];
int canget[MAX+1];

State
state(int a, int b, int c)
{
    State s;

    s.a[0] = a;
    s.a[1] = b;
    s.a[2] = c;
    return s;
}

int cap[3];

/* pour from bucket "from" to bucket "to" */
State
pour(State s, int from, int to)
{
    int amt;

    amt = s.a[from];
    if(s.a[to]+amt > cap[to])
	amt = cap[to] - s.a[to];

    s.a[from] -= amt;
    s.a[to] += amt;
    return s;
}

void
search(State s)
{
    int i, j;

    if(seen[s.a[0]][s.a[1]][s.a[2]])
	return;

    seen[s.a[0]][s.a[1]][s.a[2]] = 1;

    if(s.a[0] == 0)	/* bucket A empty */
	canget[s.a[2]] = 1;

    for(i=0; i<3; i++)
    for(j=0; j<3; j++)
	search(pour(s, i, j));	
}

void
main(void)
{
    int i;
    FILE *fin, *fout;
    char *sep;

    fin = fopen("milk3.in", "r");
    fout = fopen("milk3.out", "w");
    assert(fin != NULL && fout != NULL);

    fscanf(fin, "%d %d %d", &cap[0], &cap[1], &cap[2]);

    search(state(0, 0, cap[2]));

    sep = "";
    for(i=0; i<=cap[2]; i++) {
	if(canget[i]) {
	    fprintf(fout, "%s%d", sep, i);
	    sep = " ";
	}
    }
    fprintf(fout, "\n");

    exit(0);
}

Ran Pang from Canada sends this non-recursive DP solution:

#include<stdio.h>

int m[21][21][21];
int poss[21];
int A, B, C;

int main(void) {
    int i,j,k;
    int flag;
    FILE* in=fopen("milk3.in","r");
    fscanf(in, "%d %d %d",&A, &B, &C);
    fclose(in);
    for(i=0;i<21;i++)
        for(j=0;j<21;j++)
            for(k=0;k<21;k++)
                m[i][j][k]=0;
    for(i=0;i<21;i++)
        poss[i]=0;
    m[0][0][C]=1;

    for(flag=1;flag;) {
        flag=0;
        for(i=0;i<=A;i++)
            for(j=0;j<=B;j++)
                for(k=0;k<=C;k++) {
                    if(m[i][j][k]) {
                	if(i==0) poss[k]=1;
		        if(i) {
	                    if(j<B) {
                                if(B-j>=i) {
                            	    if( m[0][j+i][k]==0) {
                                        m[0][j+i][k]=1;
                                	flag=1;
                            	    }
                                } else {
                            	    if( m[i-(B-j)][B][k] == 0) {
                                        m[i-(B-j)][B][k] =1;
                                        flag=1;
                                    }
                                }
                            }
                            if(k<C) {
                                if(C-k>=i) {
                                    if( m[0][j][k+i]==0) {
                                        m[0][j][k+i]=1;
                                        flag=1;
                                    }
                                }
                                else {
                                    if( m[i-(C-k)][j][C] == 0) {
                                        m[i-(C-k)][j][C] =1;
                                        flag=1;
                                    }
                                }
                            }
                        }
                        if(j) {
                            if(i<A) {
                                if(A-i>=j) {
                                    if( m[i+j][0][k]==0) {
                                        m[i+j][0][k]=1;
                                        flag=1;
                                    }
                                } else {
                                    if( m[A][j-(A-i)][k] == 0) {
                                        m[A][j-(A-i)][k] =1;
                                        flag=1;
                                    }
                                }
                            }
                            if(k<C) {
                                if(C-k>=j) {
                                    if( m[i][0][k+j]==0) {
                                        m[i][0][k+j]=1;
                                        flag=1;
                                    }
                                } else {
                                    if( m[i][j-(C-k)][C] == 0) {
                                        m[i][j-(C-k)][C] =1;
                                        flag=1;
                                    }
                                }
                            }
                        }
                        if(k) {
                            if(i<A) {
                                if(A-i>=k) {
                                    if( m[i+k][j][0]==0) {
                                        m[i+k][j][0]=1;
                                        flag=1;
                                    }
                                } else {
                                    if( m[A][j][k-(A-i)] == 0) {
                                        m[A][j][k-(A-i)] =1;
                                        flag=1;
                                    }
                                }
                            }
                            if(j<B) {
                                if(B-j>=k) {
                                    if( m[i][j+k][0]==0) {
                                        m[i][j+k][0]=1;
                                        flag=1;
                                    }
                                } else {
                                    if( m[i][B][k-(B-j)] == 0) {
                                        m[i][B][k-(B-j)] =1;
                                        flag=1;
                                    }
                                }
                            }
                        }
            }                   
        }
    }
    {
        FILE* out=fopen("milk3.out", "w");
        for(i=0;i<21;i++) {
            if(poss[i]) {
                fprintf(out,"%d",i);
                i++;
                break;
            }
        }
        for(;i<21;i++) {
            if(poss[i]) {
                fprintf(out, " %d", i);
            }
        }
        fprintf(out,"\n");
    }
    return 0;
}

Daniel Jasper from Germany writes:

Both other solutions (recursive and non-recursive) use a 3D-array to store the states, so that the memory usage is O(N³). However a 2D Array and O(N²) would be enough since a state is uniquely defined by the amount of milk in bucket B and C. The amount of milk in bucket A is size-of-C minus amount-in-C minus amount-in-B. This solution works with it, and is a little bit shorter (though not more elegant):

#include <stdio.h>
int A, B, C;
int CB[21][21]; // All states

void readFile() {
    FILE *f;
    f = fopen("milk3.in", "r");
    fscanf(f, "%d%d%d", &A, &B, &C);
    fclose(f);
}

void writeFile() {
    FILE *f; int i;
    f = fopen("milk3.out", "w");
    for(i = 0; i <= C; i++) {
        if(CB[i][C - i] == 1) {
            if((i != C-B) && (i != 0)) fprintf(f, " ");
            fprintf(f, "%d", i);
        }
    }
    fprintf(f, "\n");
    fclose(f);
}

// do brute-force search, c/b: current state
void search(int c, int b) {
    int a;
    if(CB[c][b] == 1) return; // already searched
    CB[c][b] = 1;
    a = C-b-c; // calc amount in A
    // do all moves:
    // c->b
    if(B < c+b) search(c - (B - b), B);
    else search(0, c + b);
    // b->c
    if(C < c+b) search(C, b - (C - c));
    else search(c + b, 0);
    // c->a
    if(A < c+a) search(c - (A - a), b);
    else search(0, b);
    // a->c
    if(C < c+a) search(C, b);
    else search(c + a, b);
    // b->a
    if(A < b+a) search(c, b - (A - a));
    else search(c, 0);
    // a->b
    if(B < b+a) search(c, B);
    else search(c, b + a);
   }
   
int main () {
    readFile();
    search(C, 0);
    writeFile();
    return 0;
}