Codeforces 987D(BFS)

Codeforces 987D
题意：给你$n$个点$m$条边，$k$种货物，和一个$s$，之后让你从每个点开始走，收集到$s$种货物的最短路径长度。

因为$k$小只有$100$，所以我们直接把货物种类相同的用一个 vector 存起来，然后每次 BFS 每一种货物找他们到每个点的最短距离。每个点存每种货物到这个点的最短距离。之后询问排序这些距离取最小的$s$个即可。

知识点：数据小，就可以对着这个东西乱搞(暴力啊，存下来开桶什么的)

Codeforces Submission

#include<cstdio> 
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
#define db double
using namespace std;
int n, m, k, s, whw[100000 + 5][100 + 5], a[100000 + 5];
queue<int> q;
vector<int> G[100000 + 5];
vector<int> gd[100 + 5];
void ins(int x, int y) {G[x].push_back(y), G[y].push_back(x);}
void clean() {
    ms(whw, -1);
}
int solve() {
    clean();
    scanf("%d%d%d%d", &n, &m, &k, &s);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), gd[a[i]].push_back(i);
    for (int x, y, i = 1; i <= m; i++) scanf("%d%d", &x, &y), ins(x, y);
    for (int i = 1; i <= k; i++) {
        if ((int)gd[i].size() == 0) continue;
        for (int j = 0; j < (int)gd[i].size(); j++) q.push(gd[i][j]), whw[gd[i][j]][i] = 0;
        while (!q.empty()) {
            int u = q.front(); q.pop();
            for (int k = 0; k < (int)G[u].size(); k++) {
                int v = G[u][k];
                if (whw[v][i] < 0) whw[v][i] = whw[u][i] + 1, q.push(v);
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        sort(whw[i] + 1, whw[i] + 1 + k);
        int ans = 0;
        for (int j = 1; j <= s; j++) ans += whw[i][j];
        printf("%d ", ans);
    }
    return 0; 
}
int main() {
    solve();
    return 0;
}