Dancing Cows 训练赛题目--低级匹配（暴力） [SPOJ-DCOWS]

题目链接
Dancing Cows SPOJ - DCOWS

It’s the spring dance and, in a rare occurrence, the N (1 ≤ N ≤ 5000) bulls have been invited to dance with the M (N < M ≤ 5000) cows (as long as they stay on their very best behavior).

Farmer John, almost obsessive-compulsive in his organization of dances, wants the spectacle to be as visually attractive as possible. Thus, he wants to pair up the N bulls with cow partners such that the total of all the magnitudes of differences in height is minimized. Bulls have heights B_i (1 ≤ B_i ≤ 1,000,000) and cows have height C_i (1 ≤ C_i ≤ 1,000,000). Of course, some cows will be unmatched since N-M of them will have no partners; ignore their heights.

INPUT FORMAT:

• Line 1: Two space-separated integers: N and M.
• Lines 2..N+1: Line i+1 contains a single integer: B_i.

• Lines N+2..M+N+1: Line i+N+1 contains a single integer: C_i.
  OUTPUT FORMAT:

• Line 1: A single integer that is the minimum of the sum of the absolute value of the height differences that can be achieved.

SAMPLE INPUT :

5 7
10
16
12
10
13
7
17
12
10
9
6
11
SAMPLE OUTPUT :

4
INPUT DETAILS:

Five bulls + seven cows with various heights:
Bulls: 10 10 12 13 16
Cows: 6 7 9 10 11 12 17
OUTPUT DETAILS :

Here is one way to achieve a total difference of 4:
Bulls: 10 10 12 13 16
Cows: 6 7 9 10 11 12 17

题意：n只公牛，m只母牛，使公牛和母牛完成匹配，使得匹配的母牛和公牛其差值的绝对值之和最小
思路：dfs暴力匹配
code:
样例

5 710 10 12 13 166 7 9 10 11 12 17   

b: 0 c: 0b: 1 c: 1b: 2 c: 2b: 3 c: 3b: 4 c: 4b: 5 c: 50 16 11 5b: 4 c: 5b: 5 c: 60 16 12 4b: 4 c: 6b: 5 c: 70 16 17 1b: 4 c: 7b: 4 c: 7*b: 4 c: 6*b: 4 c: 5*1 13 10 3b: 3 c: 4b: 4 c: 51 13 11 2b: 3 c: 5b: 4 c: 61 13 12 1b: 3 c: 6b: 4 c: 71061109567 13 17 4b: 3 c: 7b: 3 c: 7*b: 3 c: 6*b: 3 c: 5*b: 3 c: 4*2 12 9 3b: 2 c: 3b: 3 c: 42 12 10 2b: 2 c: 4b: 3 c: 52 12 11 1b: 2 c: 5b: 3 c: 61061109567 12 12 0b: 2 c: 6b: 3 c: 71061109567 12 17 5b: 2 c: 7b: 2 c: 7*b: 2 c: 6*b: 2 c: 5*b: 2 c: 4*b: 2 c: 3*3 10 7 3b: 1 c: 2b: 2 c: 33 10 9 1b: 1 c: 3b: 2 c: 43 10 10 0b: 1 c: 4b: 2 c: 51061109567 10 11 1b: 1 c: 5b: 2 c: 61061109567 10 12 2b: 1 c: 6b: 2 c: 71061109567 10 17 7b: 1 c: 7b: 1 c: 7*b: 1 c: 6*b: 1 c: 5*b: 1 c: 4*b: 1 c: 3*b: 1 c: 2*3 10 6 4b: 0 c: 1b: 1 c: 23 10 7 3b: 0 c: 2b: 1 c: 33 10 9 1b: 0 c: 3b: 1 c: 41061109567 10 10 0b: 0 c: 4b: 1 c: 51061109567 10 11 1b: 0 c: 5b: 1 c: 61061109567 10 12 2b: 0 c: 6b: 1 c: 71061109567 10 17 7b: 0 c: 7b: 0 c: 7*b: 0 c: 6*b: 0 c: 5*b: 0 c: 4*b: 0 c: 3*b: 0 c: 2*b: 0 c: 1*4

理解代码

#include<cstdio>#include<cstring>#include<cstdlib>#include<bits/stdc++.h>using namespace std;const int  maxn = 6000;const int inf = 0x3f3f3f3f;long long dp[maxn][maxn];long long bi[maxn],ci[maxn];long long n,m;long long cal(int b,int c){    printf("b: %d c: %d\n",b,c);    if(b==n)        return 0;    if(c==m)        return inf;    if(dp[b][c]!=-1)        return dp[b][c];    long long ret = cal(b+1,c+1);    printf("%lld %lld %lld %lld\n",ret,bi[b],ci[c],abs(bi[b]-ci[c]));    ret +=abs(bi[b]-ci[c]);    long long temp = cal(b,c+1);    ret = min(ret,temp);        cout<<endl;    return dp[b][c] = ret;}int main(){    memset(dp,-1,sizeof(dp));    scanf("%lld%lld",&n,&m);    for(int i=0;i<n;i++)        scanf("%lld",&bi[i]);    for(int i=0;i<m;i++)    {        scanf("%d",&ci[i]);    }    sort(bi,bi+n);    sort(ci,ci+m);    printf("%lld\n",cal(0,0));}

code2(AC): 不要额外开很大的空间，会导致超时

#include<cstdio>#include<cstring>#include<cstdlib>#include<bits/stdc++.h>using namespace std;const int  maxn = 5010;const long long inf = 1e17;long long dp[maxn][maxn];long long bi[maxn],ci[maxn];long long n,m;long long cal(int b,int c){    //printf("b: %d c: %d\n",b,c);    if(b==n)        return 0;    if(c==m)        return inf;    if(dp[b][c]!=-1)        return dp[b][c];    long long ret = cal(b+1,c+1)+abs(bi[b]-ci[c]);  //  printf("%lld %lld %lld %lld\n",ret,bi[b],ci[c],abs(bi[b]-ci[c]));    //long long temp = cal(b,c+1);    ret = min(ret,cal(b,c+1));     //   cout<<endl;    return dp[b][c] = ret;}int main(){    memset(dp,-1,sizeof(dp));    scanf("%lld%lld",&n,&m);    for(int i=0;i<n;i++)        scanf("%lld",&bi[i]);    for(int i=0;i<m;i++)    {        scanf("%lld",&ci[i]);    }    sort(bi,bi+n);    sort(ci,ci+m);    printf("%lld\n",cal(0,0));}

dp:

#include<bits/stdc++.h>using namespace std;long long dp[5001][5001],sel;const long long maxn = 1e16;int main(){    int n,m;    scanf("%d %d",&n,&m);    int bulls[5001]= {0},cows[5001]= {0};    for(int i=1; i<=n; i++)        scanf("%d",&bulls[i]);    for(int i=1; i<=m; i++)        scanf("%d",&cows[i]);    sort(bulls,bulls+n+1);    sort(cows,cows+m+1);    for(int i=0; i<=m; i++)        dp[0][i]=0;    for(int i=1; i<=n; i++)        dp[i][0]=maxn;    for(int i=1; i<=n; i++)    {        for(int j=1; j<=m; j++)        {            dp[i][j]=min(dp[i][j-1],dp[i-1][j-1]+abs(bulls[i]-cows[j]));        }    }    printf("%lld\n",dp[n][m]);    return 0;}