Codeforces547D Mike and Fish

本文介绍了解决平面上n个点进行红蓝染色的问题,确保任意一行或一列上的红点和蓝点之差不超过1。通过两种方法实现:一种是建立图论模型,另一种是直接在x和y方向上操作并添加特定数量的边。最终通过DFS遍历完成染色。

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题意:一个平面上有n个点,先要求对每个点进行红蓝染色,且满足以下条件:对于任意一横行或一竖列,上面的红点和蓝点的个数之差最多为1。题目保证有解。

题解:出题人的解法有点厉害。。但是cubelove的做法更神。。被治愈了。。

做法1:将x坐标和y坐标分别放到两个点集中。如果有点(a,b),那么x点集中的a和y点集中的b连边。然后会有一个图。每条边都代表一个点。现在问题等价于对每个边染色,使得对于两个点集中的任意一个点所连的边的两种颜色的个数之差不大于1。如果所有点的度都为偶数,那么跑一遍欧拉回路,然后对走过的边轮流染红蓝。如果有的点的度数为奇数,那么就暂时先删掉它的某条边,然后归结于子问题,将其它的点按此种方法染色好以后,再考虑当时去的边应该染什么颜色。

做法2:考虑以下性质:如果某行或某列中有k个点,那么可以对这k个点加上[k/2]条边,使得每个点最多连了一条边。此时对每条边两端染不同的颜色就解决了。如果有孤立点,任意染色。所以我们可以这样对x方向和y方向操作,然后连边。最后对每个点dfs一下,保证每条边的两边的颜色不同,答案就出来了。



被做法2治愈了。。。


#include<bits/stdc++.h>
#define pb push_back
using namespace std;
const int N=201111;
vector<int> v[N];
int t[N],a[N][2];
void dfs(int x,int y){
    if(t[x]) return ;
    t[x]=y;
    for(int i=0;i<v[x].size();i++) dfs(v[x][i],3-y);
}
int main()
{
    int i,j,n,x,y;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        for(j=0;j<2;j++){
            scanf("%d",&x);
            if(a[x][j]){
                v[i].pb(a[x][j]);
                v[a[x][j]].pb(i);
                a[x][j]=0;
            }else a[x][j]=i;
        }
    }
    for(i=1;i<=n;i++) dfs(i,1),putchar(t[i]>1?'r':'b');
}


### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
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