[Codeforces Round 486A] Fair

[题目链接]

         https://codeforces.com/contest/986/problem/A

[算法]

        用dist(i,j)表示第i种食物运到第j个城市需要的最小代价

        将所有特产为第i中食物的城市加入队列 ， 进行广度优先搜索BFS

        显然 ， 对于每个城市 ， 答案为到该城市代价前s小的食物代价和

        时间复杂度 ： O(k(m + n))

[代码]

        

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
const int MAXK = 110;
const int inf = 1e9;

struct edge
{
        int to , nxt;
} e[MAXN << 1];

int n , m , k , s , tot;
int value[MAXN],head[MAXN];
int dist[MAXK][MAXN];
vector< int > a[MAXK];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void addedge(int u,int v)
{
        tot++;
        e[tot] = (edge){v,head[u]};
        head[u] = tot;
}

int main()
{
        
        read(n); read(m); read(k); read(s);
        for (int i = 1; i <= n; i++)
        {
                int x;
                read(x);
                a[x].push_back(i);
        }
        for (int i = 1; i <= m; i++) 
        {
                int u , v;
                read(u); read(v);
                addedge(u,v);
                addedge(v,u);
        }
        for (int i = 1; i <= k; i++) 
        {
                queue< int > q;
                while (!q.empty()) q.pop();
                for (int j = 1; j <= n; j++) dist[i][j] = inf;
                for (unsigned j = 0; j < a[i].size(); j++)
                {
                        dist[i][a[i][j]] = 0;
                        q.push(a[i][j]);        
                }        
                while (!q.empty())
                {
                        int cur = q.front();
                        q.pop();
                        for (int j = head[cur]; j; j = e[j].nxt)
                        {
                                int v = e[j].to;
                                if (dist[i][cur] + 1 < dist[i][v])
                                {
                                        dist[i][v] = dist[i][cur] + 1;
                                        q.push(v);
                                }
                        } 
                }
        }
        for (int i = 1; i <= n; i++)
        {
                for (int j = 1; j <= k; j++) value[j] = dist[j][i];
                sort(value + 1,value + k + 1);
                int ans = 0;
                for (int j = 1; j <= s; j++) ans += value[j];
                printf("%d ",ans);
        }
        printf("\n");
        
        return 0;
    
}