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  • Gym 101142C CodeCoder vs TopForces (搜索)

    题意:每个人有2种排名,对于A只要有一种排名高于B,那么A就能赢B,再如果B能赢C,那么A也能赢C,要求输出每个人分别能赢多少个人

    析:首先把题意先读对了,然后我们可以建立一个图,先按第一种排名排序,然后从高的向向低的连一条边,然后再按第二种排序,同理连线。

    最后dfs一次,要先从排名低的开始遍历,不用清0,因为是从排名低的开始的。也可以用强连通分量或者线段树。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #include <sstream>
    #define debug() puts("++++");
    #define gcd(a, b) __gcd(a, b)
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 1e5 + 10;
    const int mod = 1e9 + 7;
    const int dr[] = {-1, 0, 1, 0};
    const int dc[] = {0, 1, 0, -1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline bool is_in(int r, int c){
      return r >= 0 && r < n && c >= 0 && c < m;
    }
    vector<int> G[maxn];
    struct Node{
      int x, y, id;
    };
    Node a[maxn];
    
    bool cmp1(const Node &lhs, const Node &rhs){
      return lhs.x < rhs.x;
    }
    
    bool cmp2(const Node &lhs, const Node &rhs){
      return lhs.y < rhs.y;
    }
    int cnt;
    int ans[maxn];
    bool vis[maxn];
    
    void dfs(int u){
      if(vis[u])  return ;
      ++cnt;
      vis[u] = true;
      for(int i = 0; i < G[u].size(); ++i)  dfs(G[u][i]);
    }
    
    int main(){
      freopen("codecoder.in", "r", stdin);  
      freopen("codecoder.out", "w", stdout); 
      scanf("%d", &n);
      for(int i = 0; i < n; ++i){
        scanf("%d %d", &a[i].x, &a[i].y);
        a[i].id = i;
      }
      sort(a, a + n, cmp1);
      for(int i = n-1; i > 0; --i)  G[a[i].id].push_back(a[i-1].id);
      sort(a, a + n, cmp2);
      for(int i = n-1; i > 0; --i)  G[a[i].id].push_back(a[i-1].id);
      cnt = 0;
      for(int i = 0; i < n; ++i){
        dfs(a[i].id);
        ans[a[i].id] = cnt - 1;
      }
      for(int i = 0; i < n; ++i)  printf("%d
    ", ans[i]);
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/6628661.html
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