Tug of War
| Status |
In/Out |
TIME Limit |
MEMORY Limit |
Submit Times |
Solved Users |
JUDGE TYPE |
 |
stdin/stdout |
15s |
8192K |
479 |
114 |
Standard |
A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must not differ by more than 1; the total weight of the people on each team should be as nearly equal as possible.
The first line of input contains n the number of people at the picnic. n lines follow. The first line gives the weight of person 1; the second the weight of person 2; and so on. Each weight is an integer between 1 and 450. There are at most 100 people at the picnic.
The input may contain several test cases.
Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first.
Sample Input
3
100
90
200
Output for Sample Input
190 200
利用dp思想 ，n為偶數時求出s(n,n/2)，n為奇數時 也是s(2n,n/2)，和sum/2最接近的那個。非常經典的思路。
S(k, 1) = {A[i] | 1<= i <= k}
S(k, k) = {A[1]+A[2]+…+A[k]}
S(k, i) = S(k-1, i) U {A[k] + x | x屬于S(k-1, i-1) }
//一下代碼只能用于sum特別小的情況，否則會超時！！！！！！！！！！！
#include<iostream>
#include<cstdlib>
#define MAX 101
#define min(a,b) ((a)<(b) ? (a) : (b))
using namespace std;
int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
int num;
int a[MAX],i,j,k,l,m,NUM;
bool s[MAX][2500];
while(cin>>num)
{
int sum=0;
for(i=1;i<=num;i++)
{
cin>>a[i];
sum+=a[i];
}
if(num%2==0)NUM=num/2;
else NUM=num/2+1;
for(i=0;i<=num;i++)
for(j=0;j<=sum/2;j++)
s[i][j]=false;//表示取i個物品能否達到重量是j.
s[0][0]=true;
for(k=1;k<=num;k++)//必須在最外層，元素不能重復
for(j=min(k,NUM);j>=1; j--)//遞減的結果是使得不會出現在同一層次的互為因果 、、、、、、、、、、、巧妙的實現了課本上的序偶對生成法。
for(i=a[k];i<=sum/2;i++)
{
if(s[j-1][i-a[k]])
s[j][i]=true;
}
for(i=sum/2; i>=0; i--) {
if(s[NUM][i]) {
cout <<i<<" "<<sum-i<<endl;
break;
}
}
}
//system("PAUSE");
return 0;
}
下一次實現一個序偶生成法。
#include <iostream>
#include <functional>
using namespace std;
int a[101];
bool b[101][45002];
int main(){
// freopen("s.txt","r",stdin);
// freopen("key.txt","w",stdout);
int N,M,i,j,k;
while(scanf("%d",&N)!=EOF){
memset(b,0,sizeof(b));
a[0]=M=0;
for(i=1;i<=N;i++){
scanf("%d",a+i);
M+=a[i];
}
b[0][0]=1;
for(k=1;k<=N;k++){
for(i=1;i<=N/2;i++){
for(j=M/2;j>=0;j--){
if(b[i-1][j]){
b[i][j+a[k]]=1;
}
}
}
}
for(i=M/2,j=M/2+1;i>=0;i--,j++){
if(b[N/2][i]){
printf("%d %d\n",i,M-i);
break;
}
if(b[N/2][j]){
printf("%d %d\n",M-j,j);
break;
}
}
}
return 0;
}
posted on 2009-07-02 16:05
luis 閱讀(541)
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