题意:

杭电搬迁,有N种设备,每种设备有个价值V,数量M,要求将这些设备平分,使得平分后两边的总价值尽可能地相等。

输出两边各自的总价值。

思路:

背包DP后,P=所有的总价值/2,然后从P开始往两边找到第一个满足的价值。

可以降维,但是要注意for循环的顺序。

看代码。

代码:

int v[55], m[55];
bool dp[250005]; int main(){ int n;
while(scanf("%d",&n)!=EOF && n>=0){
int tot = 0;
rep(i,1,n){
scanf("%d%d",&v[i],&m[i]);
tot += (v[i]*m[i]);
} mem(dp,false);
dp[0]=true; rep(i,1,n){
rep(j,1,m[i]){
rep2(k,tot,j*v[i]){
dp[k]=(dp[k] || dp[k-j*v[i]]);
}
}
} int ts = tot/2;
int ans1=-1,ans2=-1;
rep2(i,ts,0){
if(dp[i] && dp[tot-i]){
ans1=i;
ans2=tot-i;
break;
}
} cout<<ans2<<" "<<ans1<<endl;
} return 0;
}

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