例题8-10(uva-714)-优快云博客

本文链接：https://blog.youkuaiyun.com/seanbill/article/details/115834665

本文介绍了一个关于物品分配的问题解决算法，通过二分查找和贪心策略确定最优分配方案。问题设定为给定一定数量的物品和限制条件，寻找将这些物品分成若干组的方法，使得每组的总和不超过给定值，并最小化组数。

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#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int maxn = 1024;
int m, k;
int input[maxn],ans[maxn];

bool Check(int value) {
	int sum = 0,cnt = 1;
	for (int i = m-1; i >=0; i--) {
		if (sum + input[i] <= value) sum += input[i];
		else {
			cnt++;
			sum = input[i];
			if (cnt > k) return false;
		}
	}
	return true;
}

int main()
{
	int t;
	cin >> t;
	while (t--) {
		cin >> m >> k;
		int nleft = 1, nRight = 0x3fffffff, nMid,sum = 0;
		memset(ans, 0, sizeof(ans));
		for (int i = 0; i < m; i++) {
			cin >> input[i];
			nleft = max(nleft, input[i]);
		}
		
		while (nleft < nRight) {
			nMid = (nleft + nRight) >> 1;
			if (Check(nMid)) {
				nRight = nMid;
			}
			else nleft = nMid + 1;
		}

		for (int i = m-1; i >= 0; i--) {
			if (sum + input[i] > nleft || i + 1 < k) { 
				k--;
				ans[i] = 1;
				sum = input[i];
			}
			else sum += input[i];
		}
		for (int i = 0; i < m - 1; i++) {
			cout << input[i] << " ";
			if (ans[i]) cout << "/ ";
		}
		cout << input[m-1] << endl;
	}
	return 0;
}