UVA 11997 K Smallest Sums （多路归并）

题意:

求k∗k矩阵每个选一个数的kk组合的前k小和

分析:

首先对于n=2的状况，两个数组为a,b
那么我们对于每个ai可以得到
1：a[1]+b[1]<=a[1]+b[2]<=...<=a[1]+b[n]
....
n：a[n]+b[1]<=a[n]+b[2]<=...<=a[n]+b[n]
对于第k个表得到a[k]+b[k]，我们使用二元组(S，B)表示一个元素，其中S=a[k]+b[k]，B是b的下标
a的下标不用记录，因为对于下一个二元组(S′,B+1)，有S‘=a[k]+b[k+1]=S−b[k]+b[k+1]
不过本题有n个表，我们可以通过两两合并就可以得出答案

代码:

////  Created by TaoSama on 2015-11-04//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, a[755], b[755];typedef pair<int, int> P;void merge() {    priority_queue<P, vector<P>, greater<P> > q;    for(int i = 1; i <= n; ++i) q.push(P(a[i] + b[1], 1));    for(int i = 1; i <= n; ++i) {        P u = q.top(); q.pop();        a[i] = u.first;        int idx = u.second;        if(idx < n)            q.push(P(u.first - b[idx] + b[idx + 1], idx + 1));    }}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%d", &n) == 1) {        for(int i = 1; i <= n; ++i) scanf("%d", a + i);        sort(a + 1, a + 1 + n);        for(int i = 2; i <= n; ++i) {            for(int j = 1; j <= n; ++j) scanf("%d", b + j);            sort(b + 1, b + 1 + n);            merge();        }        for(int i = 1; i <= n; ++i)            printf("%d%c", a[i], " \n"[i == n]);    }    return 0;}