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  • E. Turn Off The TV Educational Codeforces Round 29

    http://codeforces.com/contest/863/problem/E

    注意细节

     1 #include <cstdio>
     2 #include <cstdlib>
     3 #include <cmath>
     4 #include <cstring>
     5 #include <time.h>
     6 #include <string>
     7 #include <set>
     8 #include <map>
     9 #include <list>
    10 #include <stack>
    11 #include <queue>
    12 #include <vector>
    13 #include <bitset>
    14 #include <ext/rope>
    15 #include <algorithm>
    16 #include <iostream>
    17 using namespace std;
    18 #define ll long long
    19 #define minv 1e-6
    20 #define inf 1e9
    21 #define pi 3.1415926536
    22 #define E  2.7182818284
    23 const ll mod=1e9+7;//998244353
    24 const int maxn=2e5+10;
    25 
    26 struct node
    27 {
    28     int x,y,z;
    29 }f[maxn];
    30 
    31 bool vis[maxn]={0};
    32 
    33 int cmp(node a,node b)
    34 {
    35     if (a.x==b.x)
    36         return a.y>b.y;
    37     else
    38         return a.x<b.x;
    39 }
    40 
    41 int main()
    42 {
    43     int n,i,j,yy;
    44     scanf("%d",&n);
    45     for (i=1;i<=n;i++)
    46         scanf("%d%d",&f[i].x,&f[i].y),f[i].z=i;
    47     sort(f+1,f+n+1,cmp);
    48     vis[f[1].z]=1;
    49     j=1;
    50     for (i=1;i<=n;)
    51     {
    52         yy=f[j].y+1;
    53         i++;
    54         j=i;
    55         while (i<=n && f[i].x<=yy)
    56         {
    57             if (f[i].y>f[j].y)
    58                 j=i;
    59             i++;
    60         }
    61         if (f[j].y<=yy-1)
    62             j=i;
    63         else
    64             i--;
    65         vis[f[j].z]=1;
    66     }
    67     for (i=1;i<=n;i++)
    68         if (!vis[i])
    69         {
    70             printf("%d",i);
    71             return 0;
    72         }
    73     printf("-1");
    74     return 0;
    75 }
    76 /*
    77 3
    78 
    79 3687 7959
    80 
    81 4918 9822
    82 
    83 8049 11728
    84 */
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  • 原文地址:https://www.cnblogs.com/cmyg/p/9521050.html
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