Tug of War
| Status | In/Out | TIME Limit | MEMORY Limit | Submit Times | Solved Users | JUDGE TYPE |
|---|---|---|---|---|---|---|
| stdin/stdout | 15s | 8192K | 479 | 114 | Standard |
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为偶数时求出s(n,n/2)，n为奇数时 也是s(2n,n/2)，和sum/2最接近的那个。非常经典的思路。
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属于S(k-1, i-1) }
//一下代码只能用于sum特别小的情况，否则会超时！！！！！！！！！！！
#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个物品能否达到重量是j.
s[0][0]=true;
for(k=1;k<=num;k++)//必须在最外层，元素不能重复
for(j=min(k,NUM);j>=1; j--)//递减的结果是使得不会出现在同一层次的互为因果 、、、、、、、、、、、巧妙的实现了课本上的序偶对生成法。
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;
}
下一次实现一个序偶生成法。
#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;
}