http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3148
虽然是水题,需要用到优先队列...但还是考一点思维
题意是找出n*n的矩阵中,每行中取一个数求sum的前k小值...
可以很容易想到是求两行中的前k小然后合并...但是求两行中前k小却卡了很久...
一开始暴力把两行n^2和全部加到优先队列...直接T了...想一想O(n^3logn)确实有点过分...
实际上优先队列中只需要维护k个值就可以了...
假设两个有序数列分别为A,B ...我们先将sum{Ai,B0}入队列...
则任意一个sum{Ai,Bi+1} >= sum{Ai,Bi} 且 sum{Ai,Bi+1}不属于优先队列的集合...
对于每一个最小匹配对的sum{Ai,Bi},每次加入sum{Ai,Bi+1},就行了...
这样总的复杂度是O(n^2logn)
/********************* Template ************************/ #include <set> #include <map> #include <list> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <bitset> #include <cstdio> #include <string> #include <vector> #include <cassert> #include <cstdlib> #inclue <cstring> #include <sstream> #include <fstream> #include <numeric> #include <iomanip> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define EPS 1e-8 #define DINF 1e15 #define MAXN 805 #define LINF 1LL << 60 #define MOD 1000000007 #define INF 0x7fffffff #define PI 3.14159265358979323846 #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define BUG cout<<" BUG! "<<endl; #define LINE cout<<" ------------------ "<<endl; #define FIN freopen("in.txt","r",stdin); #define FOUT freopen("out.txt","w",stdout); #define mem(a,b) memset(a,b,sizeof(a)) #define FOR(i,a,b) for(int i = a ; i < b ; i++) #define read(a) scanf("%d",&a) #define read2(a,b) scanf("%d%d",&a,&b) #define read3(a,b,c) scanf("%d%d%d",&a,&b,&c) #define write(a) printf("%d ",a) #define write2(a,b) printf("%d %d ",a,b) #define write3(a,b,c) printf("%d %d %d ",a,b,c) #pragma comment (linker,"/STACK:102400000,102400000") template<class T> inline T L(T a) {return (a << 1);} template<class T> inline T R(T a) {return (a << 1 | 1);} template<class T> inline T lowbit(T a) {return (a & -a);} template<class T> inline T Mid(T a,T b) {return ((a + b) >> 1);} template<class T> inline T gcd(T a,T b) {return b ? gcd(b,a%b) : a;} template<class T> inline T lcm(T a,T b) {return a / gcd(a,b) * b;} template<class T> inline T Min(T a,T b) {return a < b ? a : b;} template<class T> inline T Max(T a,T b) {return a > b ? a : b;} template<class T> inline T Min(T a,T b,T c) {return min(min(a,b),c);} template<class T> inline T Max(T a,T b,T c) {return max(max(a,b),c);} template<class T> inline T Min(T a,T b,T c,T d) {return min(min(a,b),min(c,d));} template<class T> inline T Max(T a,T b,T c,T d) {return max(max(a,b),max(c,d));} template<class T> inline T exGCD(T a, T b, T &x, T &y){ if(!b) return x = 1,y = 0,a; T res = exGCD(b,a%b,x,y),tmp = x; x = y,y = tmp - (a / b) * y; return res; } template<class T> inline T reverse_bits(T x){ x = (x >> 1 & 0x55555555) | ((x << 1) & 0xaaaaaaaa); x = ((x >> 2) & 0x33333333) | ((x << 2) & 0xcccccccc); x = (x >> 4 & 0x0f0f0f0f) | ((x << 4) & 0xf0f0f0f0); x = ((x >> 8) & 0x00ff00ff) | ((x << 8) & 0xff00ff00); x = (x >>16 & 0x0000ffff) | ((x <<16) & 0xffff0000); return x; } //typedef long long LL; typedef unsigned long long ULL; //typedef __int64 LL; typedef unsigned __int64 ULL; /********************* By F *********************/ int a[MAXN][MAXN]; int n; struct node{ int val; int pos; node(int _val = 0,int _pos = 0):val(_val),pos(_pos){}; bool operator < (const node & a) const{ return val > a.val; } }; int main(){ //FIN; //FOUT; while(~scanf("%d",&n)){ for(int i = 0 ; i < n ; i++){ for(int j = 0 ; j < n ; j++) scanf("%d",&a[i][j]); sort(a[i],a[i]+n); } for(int i = 1 ; i < n ; i++){ priority_queue<node> q; int cnt = 0; for(int j = 0 ; j < n ; j++) q.push(node{a[0][j] + a[i][0],0}); while(cnt != n){ node t = q.top(); q.pop(); a[0][cnt++] = t.val; q.push(node{t.val + a[i][t.pos+1] - a[i][t.pos],t.pos+1}); } } for(int i = 0 ; i < n ; i++) printf("%d%c",a[0][i], i != n-1 ? ' ' : ' '); } return 0; }