实验12图的m着色问题

问题：

给定无向连通图$G$和$m$种颜色，用这些颜色给图的顶点着色，每个顶点一种颜色。如果要求$G$的每条边的两个顶点着不同颜色。给出所有可能的着色方案；如果不存在，则回答$”NO”$。

解析：

对于每个点选择颜色$i$时判断其与所有相连的已着色的点是否发生颜色冲突，若未发生冲突，则$dfs$向后接着选择下一个点的颜色。如果所有的点都已着色，则输出着色方案。

设计（核心代码）：

void dfs(int x) {
    if (x > n) {
        cout << ++ans << " : ";
        for (int i = 1; i <= n; ++i) {
            cout << vis[i] << " ";
        }
        cout << endl;
        return;
    }
    for (int i = 1; i <= m; ++i) {
        vis[x] = i;
        if (check(x, i))    dfs(x + 1);
        vis[x] = 0;
    }
}

分析：

复杂度：$O(n*m^{n})$。

源码：

https://github.com/Big-Kelly/Algorithm

//#include<bits/stdc++.h>
#include<time.h>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>

#define ll long long
#define pll pair<ll,ll>
#define pii pair<int,int>
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define ls root<<1
#define rs root<<1|1
#define pb push_back
#define PI acos(-1.0)

using namespace std;
const int inf = 0x3f3f3f3f;
const ll INF = 1e18 + 7;
const int maxn = 100 + 5;
const ll mod = 1e9 + 7;
const double eps = 1e-6;

inline ll read() {
    bool f = 0;
    ll x = 0; char ch = getchar();
    while (ch < '0' || ch>'9') { if (ch == '-')f = 1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar(); }
    return f ? -x : x;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a % b);
}

int n, m, cnt;
int mp[maxn][maxn], vis[maxn], ans;

void init() {
    memset(mp, 0, sizeof mp);
    memset(vis, 0, sizeof vis);
    ans = 0;
}

bool check(int x, int col) {
    for (int i = 1; i <= n; ++i) {
        if (mp[x][i]) {
            if (vis[i] == col) return 0;
        }
    }
    return 1;
}

void dfs(int x) {
    if (x > n) {
        cout << ++ans << " : ";
        for (int i = 1; i <= n; ++i) {
            cout << vis[i] << " ";
        }
        cout << endl;
        return;
    }
    for (int i = 1; i <= m; ++i) {
        vis[x] = i;
        if (check(x, i))    dfs(x + 1);
        vis[x] = 0;
    }
}

int main() {
    init();//初始化
    cin >> n >> m >> cnt;//输入点数，颜色数，边数
    while (cnt--) {
        int u, v;
        cin >> u >> v;
        mp[u][v] = mp[v][u] = 1;//建立关系
    }
    dfs(1);
    if (!ans) {
        cout << "NO" << endl;
    }
}