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  • Codeforces 986A. Fair(对物品bfs暴力求解)

    解题思路:

      1.对物品i bfs,更新每个小镇j获得每个物品i的最短距离。

      2.时间复杂度o(n*k),满足2s的要求。

    代码:

    #include <iostream>
    #include <queue>
    #include <list>
    #include <algorithm>
    #include <stdio.h>
    #include <string.h>
    using namespace std;
    typedef long long ll;
    #define MAX 2000000000 
    
    int a[500050];
    list <int> e[500050];
    int d[500050][110];
    bool used[500010];
    queue <int> q;
    
    
    int main(){
    //    fill(d[0],d[0]+500050*110,MAX);
        int n,m,k,s;
        scanf("%d%d%d%d", &n, &m, &k, &s);
        for(int i = 1;i <= n; ++i){
            scanf("%d", &a[i]);
        }
        int u,v;
        for(int i = 1;i <= m; ++i){
            scanf("%d%d", &u, &v);
            e[u].push_back(v);
            e[v].push_back(u);
        }
        for(int i = 1;i <= k; ++i){
            memset(used,false,sizeof(used));
            for(int j = 1;j <= n; ++j){
                if(a[j] == i){
                    d[j][i] = 0;
                    q.push(j);
                    used[j] = true;
                }
            }
            while(q.size()){
                int x = q.front();
                for(auto l:e[x]){
                    if(!used[l]){
                        d[l][i] = d[x][i]+1;
                        used[l] = true;
                        q.push(l);
                    }
                }
                q.pop();
            }
        }
        for(int i = 1;i <= n; ++i){
            sort(d[i]+1,d[i]+1+k);
            int ans = 0;
            for(int j = 1;j <= s; ++j){
                ans += d[i][j];
            }
            printf("%d ",ans);
        }
        printf("
    ");
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zhangjiuding/p/9112273.html
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