链接：

https://vjudge.net/problem/POJ-1129#author=SWUN2018

题意：

用大写字母表示一个广播电台，用数字代表广播电台的无线频谱，相邻两个电台之间如果运用同样的无线频谱就会互相干扰，所以我们不会让两个相邻电台之间使用同样的无线频谱.
现在问你最少使用几种无线频谱使得这些连接的广播电台之间能够正常运作？

思路：

每个点染色，保证相邻的两个点颜色不同

代码：

//#include<bits/stdc++.h>
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<string.h>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<math.h>
#include<stdio.h>
#include<map>
#include<stack>
#include<list>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int MOD = 20071027;
const int MAXN = 1010;
const double eps = 1e-3;
int Next[4][2] = {-1, 0, 0, 1, 1, 0, 0, -1};

int n, m;
vector<int> G[30];
int color[30], vis[30];
char s[50];

int solve()
{
    memset(color, -1, sizeof(color));
    for (int i = 0;i < n;i++)
    {
        memset(vis, 0, sizeof(vis));
        for (int j = 0;j < (int)G[i].size();j++) if (color[G[i][j]] != -1)
            vis[color[G[i][j]]] = 1;
        for (int c = 1;c <= i+1;c++) if (!vis[c])
        {
            color[i] = c;
            break;
        }
    }
    int ans = 0;
    for (int i = 0;i < n;i++) ans = max(ans, color[i]);
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    while(~scanf("%d", &n) && n)
    {
        for (int i = 1;i <= n;i++) G[i].clear();
        for (int i = 1;i <= n;i++)
        {
            scanf("%s", s);
            int len = strlen(s);
            for (int j = 2;j < len;j++)
                G[s[0]-'A'].push_back(s[j]-'A');
        }
        int ans = solve();
        if (ans > 1)
            printf("%d channels needed.\n", ans);
        else
            printf("%d channel needed.\n", ans);
    }

    return 0;
}

标签：频谱,1129,vis,int,long,Allocation,color,无向,include
来源： https://www.cnblogs.com/YDDDD/p/12245851.html