poj1014 Dividing

某书上的代码

#include<stdio.h>
#include<string.h>
int a[7];
void dp(int sum){
int i,j,k;
int mid=sum/2;
char vis[200];
memset(vis,0,sizeof(vis));
int t;
vis[0]=1;
for(i=1;i<=6;i++)
for(j=mid;j>=0;j--)
if(vis[j])
for(k=1;k<=a[i];k++){
t=j+i*k;
if(t>mid) break;
else vis[t]=1;
if(t==mid){
printf("Can be divided.\n\n");
return ;
}
}
printf("Can't be divided.\n\n");
}
int main(){
int i;
int icase=1;
while(1){
int sum=0;
for(i=1;i<=6;i++){
scanf("%d",&a[i]);
if(a[i]!=0&&a[i]%6==0) a[i]=6;
else a[i]%=6;
sum+=a[i]*i;
}
if(sum==0) break;
printf("Collection #%d:\n",icase++);
if(sum%2){
printf("Can't be divided.\n\n");
continue;
}
dp(sum);
}
return 0;
}

某神套用背包模板的方法，该模板来自背包九讲

#include<iostream>
#include<cstring>
#include<cstdio>
#define max(a,b) a>b?a:b
using namespace std;
int n[7];
int v;
int sumvalue;
bool flag;
int dp[100000];
void completepack(int cost,int weight){
for(int i=cost;i<=v;i++){
dp[i]=max(dp[i],dp[i-cost]+weight);
if(dp[i]==v){
flag=true;
return;
}
}
return;
}
void zeroonepack(int cost,int weight){
for(int i=v;i>=cost;i--){
dp[i]=max(dp[i],dp[i-cost]+weight);
if(dp[i]==v){
flag=true;
return;
}
}
return;
}
void multiplepack(int cost,int weight,int amount){
if(cost*amount>=v){
completepack(cost,weight);
return;
}
if(flag)
return;
int k=1;
while(k<amount){
zeroonepack(k*cost,k*weight);
if(flag)
return;
amount-=k;
k*=2;
}
zeroonepack(amount*cost,amount*weight);
return;
}
int main(){
int i;
int test=1;
while(scanf("%d%d%d%d%d%d",&n[1],&n[2],&n[3],&n[4],&n[5],&n[6])){
sumvalue=0;
for(i=1;i<=6;i++){
if(n[i]!=0&&n[i]%6==0) n[i]=6;
else n[i]=n[i]%6;
sumvalue+=i*n[i];
}
if(sumvalue==0)
break;
if(sumvalue%2){
printf("Collection #%d:\n",test++);
printf("Can't be divided.\n\n");
continue;
}
v=sumvalue/2;
memset(dp,-1,sizeof(dp));
dp[0]=0;
flag=false;
for(i=1;i<=6;i++){
multiplepack(i,i,n[i]);
if(flag)
break;
}
if(flag){
printf("Collection #%d:\n",test++);
printf("Can be divided.\n\n");
continue;
}
else{
printf("Collection #%d:\n",test++);
printf("Can't be divided.\n\n");
continue;
}
}
return 0;
}

该神还用dfs() A过去了，贪心性质的dfs()，并没有枚举所有状态，只枚举了极少状态，先用价值最大的填充，价值大的用光了或装不进去了就换次大的，赤裸裸的贪心啊，一定没有回溯，否则会tle,至于为什么可以这样做...

#include<iostream>
using namespace std;

int n[7];  //价值为i的物品的个数
int SumValue;  //物品总价值
int HalfValue;  //物品平分价值
bool flag;    //标记是否能平分SumValue

void DFS(int value,int pre)
{
if(flag)
return;

if(value==HalfValue)
{
flag=true;
return;
}

for(int i=pre;i>=1;i--)
{
if(n[i])
{
if(value+i<=HalfValue)
{
n[i]--;
DFS(value+i,i);

if(flag)
break;
}
}
}
return;
}

int main(int i)
{
int test=1;
while(cin>>n[1]>>n[2]>>n[3]>>n[4]>>n[5]>>n[6])
{
SumValue=0;  //物品总价值

for(i=1;i<=6;i++)
SumValue+=i*n[i];

if(SumValue==0)
break;

if(SumValue%2)    //sum为奇数，无法平分
{
cout<<"Collection #"<<test++<<':'<<endl;
cout<<"Can't be divided."<<endl<<endl;    //注意有空行
continue;
}

HalfValue=SumValue/2;
flag=false;

DFS(0,6);

if(flag)
{
cout<<"Collection #"<<test++<<':'<<endl;
cout<<"Can be divided."<<endl<<endl;
continue;
}
else
{
cout<<"Collection #"<<test++<<':'<<endl;
cout<<"Can't be divided."<<endl<<endl;
continue;
}
}
return 0;
}

上两种方法转自/article/1968966.html