芽衣：染上你的颜色

染色初步...

给图上的顶点指定颜色，使图上任何临接的顶点都被染成不同的颜色

我们想要知道的是最少使用几种颜色以及具体方案

《组合数学》上面有个贪心的染色法...

设G是图，它的顶点按某一顺序记为x1，x2, ..., xn。

（1）对顶点x1指定颜色1

（2）对每个i=2，3，...，n，令p是不与xi邻接的顶点的颜色重复的, (序号)最小的颜色，对xi指定颜色p

 

Problem H. Path or Coloring

Input file: Output file: Time limit: Memory limit:

pathorcoloring.in pathorcoloring.out 1 second
256 mebibytes

Do you like NP-hard problems? For example, let us consider the following two:

Coloring problem. You are given an undirected graph G and an integer k. For each vertex of the graph, choose color φ(v) so that all colors are integers between 1 and k, and any two vertices connected by an edge have different color.

Simple k-path problem. You are given an undirected graph G and an integer k. Find a simple path of length k. Formally, you need to choose such sequence of k + 1 unique vertices v1, v2, . . . , vk+1 that any two consecutive verices are connected by an edge.

You are given an undirected graph G and an integer k. Solve any one of these two problems for them. The choice is yours.

Input

The first line of input contains an integer T , the number of test cases (1 ≤ T ≤ 1000). The description of each test case consists of several lines.

The first line of each description of containts two integers n and m: the number of vertices and the number of edges in the graph (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10 000). On the next line, the integer k is given (1 ≤ k ≤ n). Each of the next m lines contains two integers a and b: the numbers of two verices connected by an edge (1 ≤ a, b ≤ n).

It is guaranteed that the given graphs have no self-loops and no multiple edges. Additionally, the total number of edges in all given graphs does not exceed 100 000.

Output

For each test case, print one of the following:

• The word “path” followed by k + 1 numbers of vertices: a simple path of length k.
• The word “coloring” followed by n integers from 1 to k: a coloring of the graph into k colors. • The word “neither” if there is no simple path of length k and no coloring into k colors.

If there exist both a valid path and a valid coloring, you can print either a path or a coloring. If there are several valid paths or several valid colorings, you can print any one path or any one coloring, respectively.

尼玛啊我不喜欢NPhard问题！

给出一个简单图，和一个数字k，要求任意选一个问题做答：

1.输出图的一个k染色方案

2.输出图的一个长度为k的简单路径

如果两种都无解输出neither

 

不知道怎么做k长简单路径，就g了。实际上：

如果有k染色方案，输出即可

如果没有k染色方案，一定存在k长简单路径，neither不存在的

因为染色数小于等于最长路径：

考虑完全图Kn，染色数为n，最长路径n。

任意删去一些边（保持连通），染色数会减小或不变

 

如果在贪心染色时，在给随意一个点指定颜色1后，从这个点邻接的点进行接下来的染色（dfs序）

那这个路径就是：颜色1，2, ..., k, k+1对应的顶点

#include
const int maxn = 1e3+7, maxm = 2e4+7;
int v[maxm], col[maxn], vis[maxn], head[maxn], nxt[maxm];
int n, m, k, tot, pos;
void addedge(int x, int y){
    v[++tot] = y;
    nxt[tot] = head[x];
    head[x] = tot;
}
void dfs(int u){
    pos++;
    for(int i = head[u]; i; i = nxt[i]){
        if(col[v[i]]){
            //printf("pos : %d ", pos);
            //printf("near %d is %d\n", u, v[i]);
            vis[col[v[i]]] = pos;
        }
    }
    for(int i = 1; ; i++){
        if(vis[i] < pos){
            col[u] = i;
            //printf("pos : %d ", pos);
            //printf("give %d color %d\n", u, i);
            break;
        }
    }
    for(int i = head[u]; i; i = nxt[i]){
        //printf("%d %d %d col v i %d\n", pos, i, v[i], col[v[i]]);
        if(!col[v[i]])
            dfs(v[i]);
    }
}
void solve(){
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 1; i <= n; i++)
        col[i] = head[i] = vis[i] = 0;
    pos = tot = 0;
    while(m--){
        int x, y;
        scanf("%d%d", &x, &y);
        addedge(x, y);
        addedge(y, x);
    }
    for(int i = 1; i <= n; i++){
        if(!col[i])
            dfs(i);
    }
    int x = 0;
    bool flag = 1;
    for(int i = 1; i <= n; i++){
        if(col[i] > k){
            x = i;
            flag = 0;
            break;
        }
    }
    /*for(int i = 1; i <= n; i++)
        printf(" %d", col[i]);
    puts("color");*/
    if(flag){
        printf("coloring");
        for(int i = 1; i <= n; i++)
            printf(" %d", col[i]);
        return;
    }
    printf("path");
    int loop = k;
    printf(" %d", x);
    while(loop){
        for(int i = head[x]; i; i = nxt[i]){
            if(col[v[i]] == loop){
                x = v[i];
                break;
            }
        }
        printf(" %d", x);
        loop--;
    }
}
int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        solve();
        puts("");
    }
    return 0;
}
/*
 1
 5 10 4
 1 2 2 3 3 4 4 5 5 1 1 3 1 4 2 5 4 2 3 5
*/

View Code

 

转载于:https://www.cnblogs.com/DearDongchen/p/7635784.html