AtCoder2362 - Splatter Painting

1.题目描述：

B - Splatter Painting

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.

Sample Input 2

Copy

Sample Output 2

Copy

The given graph may not be connected.

2.题意概述：

给一个包含N个顶点，M条边，无自环和重边的简单无向图，初始每个点颜色都为0，每条边的长度为1,连接着ai,bi两个节点。经过若干个操作，

每次将与某个点vi距离不超过di的所有点染成某种颜色ci，求最终每个点的颜色。

3.解题思路：

反过来做，染色过的点就不再染色
对于点u来说，如果以它为中心，距离为d的所有点都被染色过，那么下次你要对u距离为d’ (d’<d) 的点染色时就可以直接退出了

4.AC代码：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define maxn 100100
#define lson root << 1
#define rson root << 1 | 1
#define lent (t[root].r - t[root].l + 1)
#define lenl (t[lson].r - t[lson].l + 1)
#define lenr (t[rson].r - t[rson].l + 1)
#define N 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
vector<int> g[maxn];
int vis[maxn], col[maxn];
tuple<int, int, int> op[maxn];
void dfs(int v, int d, int c)
{if (vis[v] >= d)return;vis[v] = d;if (!col[v])col[v] = c;if (d){int sz = g[v].size();for (int i = 0; i < sz; i++)dfs(g[v][i], d - 1, c);}
}
int main()
{int n, m, q;while (~scanf("%d%d", &n, &m)){for (int i = 1; i <= n; i++)g[i].clear();fill(vis, vis + n + 1, -1);fill(col, col + n + 1, 0);while (m--){int u, v;scanf("%d%d", &u, &v);g[u].push_back(v);g[v].push_back(u);}scanf("%d", &q);for (int i = 0; i < q; i++){int v, d, c;scanf("%d%d%d", &v, &d, &c);op[i] = tuple<int, int, int>(v, d, c);}for (int i = q - 1; i >= 0; i--){int v, d, c;tie(v, d, c) = op[i];dfs(v, d, c);}for (int i = 1; i <= n; i++)printf("%d\n", col[i]);}return 0;
}