zoukankan      html  css  js  c++  java
  • Bridging signals(二分 二分+stl dp)

    Bridging signals

    Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 941    Accepted Submission(s): 614


    Problem Description
    'Oh no, they've done it again', cries the chief designer at the Waferland chip factory. Once more the routing designers have screwed up completely, making the signals on the chip connecting the ports of two functional blocks cross each other all over the place. At this late stage of the process, it is too
    expensive to redo the routing. Instead, the engineers have to bridge the signals, using the third dimension, so that no two signals cross. However, bridging is a complicated operation, and thus it is desirable to bridge as few signals as possible. The call for a computer program that finds the maximum number of signals which may be connected on the silicon surface without rossing each other, is imminent. Bearing in mind that there may be housands of signal ports at the boundary of a functional block, the problem asks quite a lot of the programmer. Are you up to the task?

    Figure 1. To the left: The two blocks' ports and their signal mapping (4,2,6,3,1,5). To the right: At most three signals may be routed on the silicon surface without crossing each other. The dashed signals must be bridged.

    A typical situation is schematically depicted in figure 1. The ports of the two functional blocks are numbered from 1 to p, from top to bottom. The signal mapping is described by a permutation of the numbers 1 to p in the form of a list of p unique numbers in the range 1 to p, in which the i:th number pecifies which port on the right side should be connected to the i:th port on the left side.
    Two signals cross if and only if the straight lines connecting the two ports of each pair do.
     

    Input
    On the first line of the input, there is a single positive integer n, telling the number of test scenarios to follow. Each test scenario begins with a line containing a single positive integer p<40000, the number of ports on the two functional blocks. Then follow p lines, describing the signal mapping: On the i:th line is the port number of the block on the right side which should be connected to the i:th port of the block on the left side.
     

    Output
    For each test scenario, output one line containing the maximum number of signals which may be routed on the silicon surface without crossing each other.
     

    Sample Input
    4 6 4 2 6 3 1 5 10 2 3 4 5 6 7 8 9 10 1 8 8 7 6 5 4 3 2 1 9 5 8 9 2 3 1 7 4 6
     

    Sample Output
    3 9 1 4
     

     题解:二分水过,dp超时,就是求递增的长度,跟最长单调子序列稍有不同,这个可以用二分,随时更新前面小的元素;

    二分:

     1 #include<stdio.h>
     2 int a[40010];
     3 int main(){
     4     int T,M,top,l,r,mid,m;
     5     scanf("%d",&T);
     6     while(T--){top=0;
     7         scanf("%d",&M);
     8         scanf("%d",&m);
     9         a[top]=m;l=0;r=top;
    10         for(int i=1;i<M;i++){l=0;r=top;mid=0;
    11             scanf("%d",&m);
    12             if(m>a[top])a[++top]=m;
    13             else{
    14              while(l<=r){
    15                 mid=(l+r)/2;
    16                 if(a[mid]>m)r=mid-1;
    17                 else l=mid+1;
    18             }
    19             a[l]=m;}
    20         }//for(int i=0;i<=top;++i)printf("%d ",a[i]);
    21         printf("%d
    ",top+1);
    22     }
    23     return 0;
    24 }

    二分+stl:

     1 #include<stdio.h>
     2 #include<algorithm>
     3 using namespace std;
     4 int a[40010];
     5 int main(){
     6     int T,N,m,top,l,r,mid;
     7     scanf("%d",&T);
     8     while(T--){top=0;
     9         scanf("%d",&N);
    10         scanf("%d",&m);
    11         a[top]=m;
    12         for(int i=1;i<N;i++){l=0;r=top;
    13             scanf("%d",&m);
    14             if(m>a[top])a[++top]=m;
    15             else *lower_bound(a,a+r,m)=m;
    16         }
    17         printf("%d
    ",top+1);
    18     }
    19     return 0;
    20 }

    dp超时:

     1 #include<stdio.h>
     2 #include<string.h>
     3 #define MAX(x,y) x>y?x:y
     4 int dp[40010];
     5 int m[40010];
     6 int main(){
     7     int T,N;
     8     scanf("%d",&T);
     9     while(T--){memset(dp,0,sizeof(dp));
    10         scanf("%d",&N);
    11         for(int i=0;i<N;++i){scanf("%d",&m[i]);dp[i]=1;
    12             for(int j=0;j<i;j++){
    13                 if(m[i]>=m[j])dp[i]=MAX(dp[j]+1,dp[i]);
    14             }
    15         }
    16         printf("%d
    ",dp[N-1]);
    17     }
    18     return 0;
    19 }
  • 相关阅读:
    Python基础 | pandas中dataframe的整合与形变(merge & reshape)
    Python基础 | pandas中数据的筛选(index & subset)
    Python基础 | 关于“循环”那些事
    Python基础 | 数据文件的读写
    Python基础 | 字符串操作
    如何用数据说谎 How to lie with data
    浅谈“数据敏感度”
    爬虫 | IT桔子互联网公司死亡名单
    从一道面试题谈数据推算方法
    爬虫 | cnblog文章收藏排行榜(“热门文摘”)
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4688795.html
Copyright © 2011-2022 走看看