Codeforces Round #485(Div. 1) 문제 해결

A. Fair
time limit per test
2 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output
Some company is going to hold a fair in Byteland. There are nn towns in Byteland and mm two-way roads between towns. Of course, you can reach any town from any other town using roads.
There are kk types of goods produced in Byteland and every town produces only one type. To hold a fair you have to bring at least ssdifferent types of goods. It costs d(u,v)d(u,v) coins to bring goods from town uu to town vv where d(u,v)d(u,v) is the length of the shortest path from uu to vv. Length of a path is the number of roads in this path.
The organizers will cover all travel expenses but they can choose the towns to bring goods from. Now they want to calculate minimum expenses to hold a fair in each of nn towns.
Input
There are 44 integers nn, mm, kk, ss in the first line of input (1≤n≤1051≤n≤105, 0≤m≤1050≤m≤105, 1≤s≤k≤min(n,100)1≤s≤k≤min(n,100)) — the number of towns, the number of roads, the number of different types of goods, the number of different types of goods necessary to hold a fair.
In the next line there are nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤k1≤ai≤k), where aiai is the type of goods produced in the ii-th town. It is guaranteed that all integers between 11 and kk occur at least once among integers aiai.
In the next mm lines roads are described. Each road is described by two integers uu vv (1≤u,v≤n1≤u,v≤n, u≠vu≠v) — the towns connected by this road. It is guaranteed that there is no more than one road between every two towns. It is guaranteed that you can go from any town to any other town via roads.
Output
Print nn numbers, the ii-th of them is the minimum number of coins you need to spend on travel expenses to hold a fair in town ii. Separate numbers with spaces.
Examples
input
Copy

5 5 4 3
1 2 4 3 2
1 2
2 3
3 4
4 1
4 5

output
Copy

2 2 2 2 3 

input
Copy

7 6 3 2
1 2 3 3 2 2 1
1 2
2 3
3 4
2 5
5 6
6 7

output
Copy

1 1 1 2 2 1 1 

Note
Let's look at the first sample.
To hold a fair in town 11 you can bring goods from towns 11 (00 coins), 22 (11 coin) and 44 (11 coin). Total numbers of coins is 22.
Town 22: Goods from towns 22 (00), 11 (11), 33 (11). Sum equals 22.
Town 33: Goods from towns 33 (00), 22 (11), 44 (11). Sum equals 22.
Town 44: Goods from towns 44 (00), 11 (11), 55 (11). Sum equals 22.
Town 55: Goods from towns 55 (00), 44 (11), 33 (22). Sum equals 33
【분석】 1<=s<=k<=min(100,n)이고 각 경로의 길이가 1이기 때문에 각 점을 기점으로 bfs를 한 번 하고 방문 종류가 s에 도달했을 때 bfs를 종료할 생각은 어렵지 않다.
쓸 때 k와 s가 작아서 매번 bfs의 입대 수량이 많지 않을 거라고 생각하고 초기화할 때 입대 수량이 많지 않기 때문에 하나의 그룹으로 수정된 원을 기록한다.
소, 끝난 후에 다시 고칩니다. 직접memset은 수조가 비교적 크기 때문에 시간을 초과할 수 있습니다.다 쓴 후에 생각해 보니 그림이 하나의 체인일 때 앞의 요소가 모두
다 똑같아요. 마지막 몇 개는 달라요. 끊을 수 있는데 A가 돼서 고치기 귀찮아요.
못 쓰게 썼으니 마음대로 보아라

#include
#include
#include
using namespace std;
int n, m, k, s;
int head[110000], ne[220000], y[220000], tot, ans;
int a[110000], dis[110000], f[11000], fb;
bool b[110000], flag[200];
int h[110000], he, ta;
void add(int u,int v)
{
    ne[++tot] = head[u];
    head[u] = tot;
    y[tot] = v;
}

void bfs(int x)
{
    he = ta = 1;
    int cnt = 1, u, p;
    h[1] = x;
    b[x] = 1;
    flag[a[x]] = 1;
    f[++fb] = x;
    while (ta >= he)
    {
        u = h[he++];
        p = head[u];
        while (p)
        {
            if (cnt >= s) return;
            if (dis[y[p]] == 0) dis[y[p]] = dis[u] + 1;
            if (!b[y[p]]) h[++ta] = y[p], b[y[p]] = 1, f[++fb] = y[p];
            if (!flag[a[y[p]]]) ++cnt, flag[a[y[p]]] = 1, ans += dis[y[p]];

            p = ne[p];
        }
    }
}
int main()
{
    int i, j, u, v;
    scanf("%d%d%d%d", &n, &m, &k, &s);
    for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (i = 1; i <= m; ++i)
    {
        scanf("%d%d", &u, &v);
        add(u, v);
        add(v, u);
    }
    for (i = 1; i <= n; ++i)
    {
        ans = 0;
        fb = 0;
        bfs(i);
        if (i != n) printf("%d ", ans);
        else printf("%d", ans);
        for (j = 1; j <= fb; ++j) dis[f[j]] = 0, flag[a[f[j]]] = 0, b[f[j]] = 0;
    }
    return 0;
}