AtCoder Grand Contest 012 B - Splatter Painting(dp)

Time limit : 2sec / Memory limit : 256MB

Score : 700 points

Problem Statement

Squid loves painting vertices in graphs.

There is a simple undirected graph consisting of N vertices numbered 1 through N, and M edges. Initially, all the vertices are painted in color 0. The i-th edge bidirectionally connects two vertices ai and bi. The length of every edge is 1.

Squid performed Q operations on this graph. In the i-th operation, he repaints all the vertices within a distance of di from vertex vi, in color ci.

Find the color of each vertex after the Q operations.

Constraints

1≤N,M,Q≤105
1≤ai,bi,vi≤N
ai≠bi
0≤di≤10
1≤ci≤105
di and ci are all integers.
There are no self-loops or multiple edges in the given graph.

Partial Score

200 points will be awarded for passing the testset satisfying 1≤N,M,Q≤2,000.

Input

Input is given from Standard Input in the following format:

N M

a1 b1

:

aM bM

Q

v1 d1 c1

:

vQ dQ cQ

Output

Print the answer in N lines. In the i-th line, print the color of vertex i after the Q operations.

Sample Input 1

Copy

Sample Output 1

Copy

Initially, each vertex is painted in color 0. In the first operation, vertices 5 and 6 are repainted in color 1. In the second operation, vertices 1, 2, 3, 4 and 5 are repainted in color 2.

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题意：给你N个点，M条边（可能有点的没有边），有Q次染色操作，每次染第vi节点的di范围内的所有节点。节点的颜色可以被覆盖。

解题思路：

1）暴力dfs

保存每一个顶点连同的顶点。

每一次输入v，d，c，搜索v节点的d范围内能够染色的点，每一次d-1;

2）dp

先贴上官方题解：https://agc012.contest.atcoder.jp/editorial;

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简单翻译一下（虽然是看着jcvb dalao理解的）

dp[i][d]表示当前i节点d距离时的颜色，所以dp[i][0]为i节点的颜色，

所以就可以在dfs时判断如果当前节点dp[i][d]已经染了色，就可以不用染色了。

还有的就是：因为颜色是覆盖的，所以最后染的颜色就是结束的颜色。所以我们应该从后往前开始染色。

这样就可以保证，如果这节点d[i][0]染了色，就不会被覆盖了。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <vector>

using namespace std;

typedef long long LL;

const int maxn = 100000+10;

int color[maxn][15];

int v[maxn];

int d[maxn];

int c[maxn];

vector <int> node[maxn];

void paint(int v1, int d1, int c1)

{

    if(d1==-1)  return;

    if(color[v1][d1] ) return;

    color[v1][d1] = c1;

    for(int i =0;i<node[v1].size();i++)

        paint(node[v1][i], d1-1, c1);

}

int main()

{

    ios::sync_with_stdio(0);

    memset(color, 0, sizeof color);

    int N, M;

    cin >> N >>M;

    for(int i=1;i<=M;i++){

        int a, b;

        cin >> a >> b;

        node[a].push_back(b);

        node[b].push_back(a);

    }

    ///自己指向自己， 才能够自己跑到color[i][0]

    for(int i=1;i<=N;i++)

        node[i].push_back(i);

    int q;

    cin >> q;

    for(int i=1;i<=q;i++)

        cin >> v[i] >> d[i] >> c[i];

    for(int i=q;i>=1;i--)

        paint(v[i], d[i], c[i]);

    for(int i=1;i<=N;i++)

        cout << color[i][0] << endl;

    return 0;

}

可以用链式前向星来代替vector。