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Published on 2017-09-02 11:31 in 暂未分类 with qscqesze

# Codeforces Round #599 (Div. 1) C. Sum Balance 图论 dp

## C. Sum Balance

Ujan has a lot of numbers in his boxes. He likes order and balance, so he decided to reorder the numbers.

There are 𝑘 boxes numbered from 1 to 𝑘. The 𝑖-th box contains 𝑛𝑖 integer numbers. The integers can be negative. All of the integers are distinct.

Ujan is lazy, so he will do the following reordering of the numbers exactly once. He will pick a single integer from each of the boxes, 𝑘 integers in total. Then he will insert the chosen numbers — one integer in each of the boxes, so that the number of integers in each box is the same as in the beginning. Note that he may also insert an integer he picked from a box back into the same box.

Ujan will be happy if the sum of the integers in each box is the same. Can he achieve this and make the boxes perfectly balanced, like all things should be?

## Input

The first line contains a single integer 𝑘 (1≤𝑘≤15), the number of boxes.

The 𝑖-th of the next 𝑘 lines first contains a single integer 𝑛𝑖 (1≤𝑛𝑖≤5000), the number of integers in box 𝑖. Then the same line contains 𝑛𝑖 integers 𝑎𝑖,1,…,𝑎𝑖,𝑛𝑖 (|𝑎𝑖,𝑗|≤109), the integers in the 𝑖-th box.

It is guaranteed that all 𝑎𝑖,𝑗 are distinct.

## Output

If Ujan cannot achieve his goal, output "No" in a single line. Otherwise in the first line output "Yes", and then output 𝑘 lines. The 𝑖-th of these lines should contain two integers 𝑐𝑖 and 𝑝𝑖. This means that Ujan should pick the integer 𝑐𝑖 from the 𝑖-th box and place it in the 𝑝𝑖-th box afterwards.

If there are multiple solutions, output any of those.

You can print each letter in any case (upper or lower).

input
4
3 1 7 4
2 3 2
2 8 5
1 10
output
Yes
7 2
2 3
5 1
10 4
input
2
2 3 -2
2 -1 5
output
No
input
2
2 -10 10
2 0 -20
output
Yes
-10 2
-20 1

## Note

In the first sample, Ujan can put the number 7 in the 2nd box, the number 2 in the 3rd box, the number 5 in the 1st box and keep the number 10 in the same 4th box. Then the boxes will contain numbers {1,5,4}, {3,7}, {8,2} and {10}. The sum in each box then is equal to 10.

In the second sample, it is not possible to pick and redistribute the numbers in the required way.

In the third sample, one can swap the numbers −20 and −10, making the sum in each box equal to −10.

## 代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 16;
vector<pair<pair<int,long long>, int>> ans[1<<maxn];
map<long long,int> color;
vector<int>a[maxn];
long long sum[maxn];
int k;
int dp[1<<maxn];
int main(){
scanf("%d",&k);
long long all_sum = 0;
for(int i=0;i<k;i++){
int n;scanf("%d",&n);
for(int j=0;j<n;j++){
long long x;scanf("%lld",&x);
a[i].push_back(x);
all_sum+=x;
sum[i]+=x;
color[a[i][j]]=i;
}
}
if(all_sum%k!=0){
puts("NO");
return 0;
}
all_sum/=k;
for(int i=0;i<k;i++){
for(int j=0;j<a[i].size();j++){
long long cur = a[i][j];
int used = 0;
bool isOk = true;
vector<pair<pair<int,long long>, int>> an;
do{
int cl = 0;
auto it = color.find(cur);
if(it!=color.end()){
cl = it->second;
}else{
isOk = false;
break;
}
if(used&(1<<cl)){
isOk = false;
break;
}
used|=(1<<cl);
cur = cur + (all_sum - sum[cl]);
auto cl2 = color.find(cur);
if(cl2!=color.end()){
an.push_back({{cl2->second,cur},cl});
}
// cout<<cur<<endl;
// cout<<cur<<" "<<cl<<" "<<cl2<<endl;
}while(cur!=a[i][j]);
if(isOk){
// cout<<"made! " << used << endl;
dp[used]=1;
ans[used] = std::move(an);
}
// cout<<"------"<<endl;
}
}

for(int i=0;i<(1<<k);i++){
if(dp[i])continue;
for(int j=i;j>0;j=(j-1)&i){
if(dp[j]&&dp[i&(~j)]){
dp[i]=1;
ans[i]=ans[j];
for(auto&x : ans[i&(~j)]){
ans[i].push_back(x);
}
break;
}
}
}
// for(int i=0;i<(1<<k);i++){
//     cout<<dp[i]<<endl;
// }
int x = (1<<k) - 1;
if (dp[x]) {
cout<<"Yes"<<endl;
sort(ans[x].begin(),ans[x].end());
for(auto a:ans[x]){
cout<<a.first.second<<" "<<a.second+1<<endl;;
}
} else {
cout<<"No"<<endl;
}
}
posted @ 2019-11-07 17:17  qscqesze  阅读(183)  评论(0编辑  收藏