URAL 1080 Map Coloring（染色）

最新推荐文章于 2022-01-18 09:07:09 发布

转载最新推荐文章于 2022-01-18 09:07:09 发布 · 300 阅读

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原文链接：http://www.cnblogs.com/jianrenfang/p/6000052.html

本文介绍了一种解决地理地图两色染色问题的算法。该算法通过深度优先搜索确定地图上的国家是否可以用红色和蓝色两种颜色进行染色，使得相邻的国家颜色不同。文章提供了一个具体的C++实现示例。

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Map Coloring

Time limit: 1.0 second
Memory limit: 64 MB

We consider a geographical map with N countries numbered from 1 to N (0 < N < 99). For every country we know the numbers of other countries which are connected with its border. From every country we can reach to any other one, eventually crossing some borders. Write a program which determines whether it is possible to color the map only in two colors — red and blue in such a way that if two countries are connected their colors are different. The color of the first country is red. Your program must output one possible coloring for the other countries, or show, that such coloring is impossible.

Input

On the first line is written the number N. On the following N lines, the i-th line contains the countries to which the i-th country is connected. Every integer on this line is bigger than i, except the last one which is 0 and marks that no more countries are listed for country i. If a line contains 0, that means that the i-th country is not connected to any other country, which number is larger than i.

Output

The output contains exactly one line. If the coloring is possible, this line must contain a list of zeros and ones, without any separators between them. The i-th digit in this sequence is the color of the i-th country. 0 corresponds to red color, and one — to blue color. If a coloring is not possible, output the integer −1.

Sample

input	output
3 2 0 3 0 0	010

Problem Author: Emil Kelevedzhiev

【分析】一个很简单的图染色问题。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
typedef long long ll;
using namespace std;
const int N = 170;
const int M = 24005;
int  n,m,k,l,tot=0;
int parent[N],pre[N],vis[N],color[N];
vector<int>vec[N];
bool ok=true;
void dfs(int u){
    vis[u]=1;
    for(int i=0;i<vec[u].size();i++){
        int v=vec[u][i];
        if(!vis[v]){
            color[v]=(color[u]+1)%2;
            dfs(v);
        }
        else if(vis[v]&&color[v]==color[u]){
            ok=false;return;
        }
    }
}
int main() {
    int u,v;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        while(1){
            scanf("%d",&v);
            if(!v)break;
            vec[i].pb(v);vec[v].pb(i);
        }
    }
    for(int i=1;i<=n;i++){
        if(!vis[i]){
            color[i]=0;
            dfs(i);
        }
    }
    if(ok)for(int i=1;i<=n;i++)printf("%d",color[i]);
    else printf("-1");
    printf("\n");

    return 0;
}

转载于:https://www.cnblogs.com/jianrenfang/p/6000052.html