AtCoder-2362 (dfs+优化)

题目

Vjudge https://vjudge.net/problem/AtCoder-2362
原网址 http://agc012.contest.atcoder.jp/tasks/agc012_b

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≤10^5
di and ci are all integers.
There are no self-loops or multiple edges in the given graph.

Input

Input is given from Standard Input in the following format:

N M
a1 b1
:
aM bM
Q
v1 d1 c1
:
vQ dQ cQ

The given graph may not be connected.

Output

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

Samples

No. Input Output 1 7 7
1 2
1 3
1 4
4 5
5 6
5 7
2 3
2
6 1 1
1 2 2 2
2
2
2
2
1
0 2 14 10
1 4
5 7
7 11
4 10
14 7
14 3
6 14
8 11
5 13
8 3
8
8 6 2
9 7 85
6 9 3
6 7 5
10 3 1
12 9 4
9 6 6
8 2 3
1
0
3
1
5
5
3
3
6
1
3
4
5
3

分析

• 题意就是给个n个点的图，m个操作，每次选择一个点v[i]把与它距离不大于d[i]的点染成颜色c[i]，问最后每个点的颜色。
• 看看数据范围（除了d[] 只有10，每个数都达到100000），直接暴力修改肯定不行。
• 可以发现，暴力之所以这么慢，是因为有许多重复染色的多余操作，那么我们把它离线化处理，从后往前，这样染过色的点就不用染（可是却还需要继续往下递归，因为不能保证下面的都是这个颜色）
• 还有一点，假如点 x 要把离它 dx 的点染色，而在这之前与它相邻的点 y已经把离它 dy 的点染过色了，这时要是 dy>=dx ，那么 x 的这一操作完全可以省去了，这样就的确快了许多。
• 可以证明一下，每个点最多被讨论 10 次（因为 d[] 最多为10），所以时间复杂度为 10n。

代码

#include <cstdio> #define For(x) for (int h=head[x],o=V[h]; h; o=V[h=to[h]])#define add(u,v) (to[++num]=head[u],head[u]=num,V[num]=v)int num,head[100005],to[200005],V[200005];int Q,n,m,v[100005],d[100005],c[100005],cl[100005],rg[100005];void dfs(int x,int ran,int col){    if (rg[x]>=ran || ran<=0) return;    rg[x]=ran;              //范围     if (!cl[x]) cl[x]=col;  //颜色     For(x) dfs(o,ran-1,col);}int main(){    freopen("1.txt","r",stdin);    scanf("%d%d",&n,&m);    for (int i=1,x,y; i<=m; i++){        scanf("%d%d",&x,&y);        add(x,y); add(y,x);    }    scanf("%d",&Q);    for (int i=1; i<=Q; i++) scanf("%d%d%d",&v[i],&d[i],&c[i]);     for (int i=Q; i; i--) dfs(v[i],d[i]+1,c[i]);    for (int i=1; i<=n; i++) printf("%d\n",cl[i]);}