11997 - K Smallest Sums(优先队列)
11997 - K Smallest Sums
You’re given k arrays, each array has k integers. There are k
k ways to pick exactly one element in each
array and calculate the sum of the integers. Your task is to find the k smallest sums among them.
Input
There will be several test cases. The first line of each case contains an integer k (2 ≤ k ≤ 750). Each of
the following k lines contains k positive integers in each array. Each of these integers does not exceed
1,000,000. The input is terminated by end-of-file (EOF).
Output
For each test case, print the k smallest sums, in ascending order.
Sample Input
3
1 8 5
9 2 5
10 7 6
2
1 1
1 2
Sample Output
9 10 12
2 2
题意:
给你k个数组,从每个数组里面取个数字,求和,让找k个最小值;
大神的优先队列,自己用好友的思路写了一遍,没考虑周全,wa了,然后借助大神的思路写了一遍就对了,大神是把复杂问题简单化,把k维转化为二维;
代码:
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=800;
typedef long long LL;
int mp[MAXN][MAXN];
struct Node{
int s,p;
Node(int s,int p):s(s),p(p){}
friend bool operator <(Node a,Node b){
return a.s>b.s;
}
};
void merge(int *a,int *b,int *c,int n){
sort(a,a+n);sort(b,b+n);
priority_queue<Node>dl;
for(int i=0;i<n;i++)dl.push(Node(a[i]+b[0],0));
for(int i=0;i<n;i++){
Node d=dl.top();dl.pop();
c[i]=d.s;
dl.push(Node(d.s-b[d.p]+b[d.p+1],d.p+1));
}
}
int main(){
int k;
while(~scanf("%d",&k)){
for(int i=0;i<k;i++)for(int j=0;j<k;j++)scanf("%d",&mp[i][j]);
for(int i=1;i<k;i++){
merge(mp[0],mp[i],mp[0],k);
}
for(int i=0;i<k;i++){
if(i)printf(" ");
printf("%d",mp[0][i]);
}
puts("");
}
return 0;}
刚开始没考虑周全的代码:
4
1 2 5 100
1 2 8 100
1 8 9 100
1 10 15 100
这组数据就没过。。
wa代码:
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=800;
typedef long long LL;
priority_queue<int,vector<int>,greater<int> >dl,q;
int main(){
int k;
while(~scanf("%d",&k)){
while(!dl.empty())dl.pop();
while(!q.empty())q.pop();
int ans=0;
int x,a,b;
for(int i=0;i<k;i++){
for(int j=0;j<k;j++){
scanf("%d",&x);dl.push(x);
}
a=dl.top();
ans+=a;
dl.pop();
b=dl.top();
q.push(b-a);
// printf("%d %d\n",a,b);
while(!dl.empty())dl.pop();
}
for(int i=0;i<k;i++){
if(i)printf(" ");
if(!i)printf("%d",ans);
else printf("%d",ans+q.top()),q.pop();
}
puts("");
}
return 0;
}
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