Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2)（574A，574B）

最新推荐文章于 2024-07-15 17:55:50 发布

原创最新推荐文章于 2024-07-15 17:55:50 发布 · 528 阅读

0 ·

CC 4.0 BY-SA版权

-------Codeforces Round----- 专栏收录该内容

21 篇文章

订阅专栏

本文提供CodeForces 574 A题Bear and Elections及B题Bear and Three Musketeers的详细解析与代码实现，介绍如何通过模拟算法解决选举问题，并提出O(n^2 + nm)复杂度的解决方案来寻找最佳三剑客组合。

1.Bear and Elections

题目链接：

http://codeforces.com/problemset/problem/574/A

解题思路：

模拟即可。。。

AC代码：

#include <iostream>
#include <cstdio>
#include <set>
#include <algorithm>
using namespace std;

multiset<int> s;
multiset<int>::iterator it;

int main(){
    int n;
    while(~scanf("%d",&n)){
        int ans,sum = 0;
        s.clear();
        scanf("%d",&ans);
        int x,tmp;
        for(int i = 2; i <= n; i++){
            scanf("%d",&x);
            s.insert(x);
        }
        it = s.end();
        it--;
        while(ans <= *it){
            sum++;
            ans++;
            tmp = (*it) - 1;
            s.erase(it);
            s.insert(tmp);
            it = s.end();
            it--;
        }
        printf("%d\n",sum);
    }
    return 0;
}

2.Bear and Three Musketeers

题目链接：

http://codeforces.com/problemset/problem/574/B

解题思路：

Warriors are vertices and "knowing each other" is an edge. We want to find connected triple of vertices with the lowest sum of

degrees (and print sum - 6 because we don't want to count edges from one chosen vertex to another).

Brute force is O(n3). We iterate over all triples a, b, c and consider them as musketeers. They must be connected by edges (they

must know each other). If they are, then we consider sum of their degrees.

We must notice that there is low limit for number of edges. So instead of iterating over triples of vertices we can iterate over edges

and then iterate over third vertex. It gives us O(n2 + nm) and it's intended solution. To check if third vertex is connected with other

two, you should additionally store edges in 2D adjacency matrix.

It's also possible to write it by adding "if" in right place in brute forces to get O(n2 + nm).

AC代码：

#include<bits/stdc++.h>
using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 5005;
int degree[maxn];
bool t[maxn][maxn];

int main(){
	int n,m;
	scanf("%d%d", &n, &m);
	for(int i = 0; i < m; ++i){
		int a, b;
		scanf("%d%d", &a, &b);
		t[a][b] = t[b][a] = true;
		degree[a]++;
		degree[b]++;
	}
	int result = INF;
	for(int i = 1; i <= n; i++)
		for(int j = i + 1; j <= n; j++){
			if(t[i][j]) {
				for(int k = j + 1; k <= n; ++k){
					if(t[i][k] && t[j][k])
						result = min(result, degree[i]+degree[j]+degree[k]);
				}
        }
    }
	if(result == INF)
        printf("-1\n");
	else
        printf("%d\n", result - 6);
	return 0;
}