UVA 111
2014-04-21 17:00
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和最长公共子序列差不多的dp
A - History Grading
Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
Submit Status Practice UVA
111
Appoint description:
System Crawler (2014-03-27)
Description
Consider a history exam in which students are asked to put several historical events into chronological order. Students who order all the events correctly will receive full credit, but how should partial credit
be awarded to students who incorrectly rank one or more of the historical events?
Some possibilities for partial credit include:
1 point for each event whose rank matches its correct rank
1 point for each event in the longest (not necessarily contiguous) sequence of events which are in the correct order relative to each other.
For example, if four events are correctly ordered 1 2 3 4 then the order 1 3 2 4 would receive a score of 2 using the first method (events 1 and 4 are correctly ranked) and a score of 3 using the second method
(event sequences 1 2 4 and 1 3 4 are both in the correct order relative to each other).
In this problem you are asked to write a program to score such questions using the second method.
as
where
denotes
the ranking of event i in the correct chronological order and a sequence of student responses
where
denotes
the chronological rank given by the student to event i; determine the length of the longest (not necessarily contiguous) sequence of events in the student responses that are in the correct chronological order relative to each other.
.
The second line will contain nintegers, indicating the correct chronological order of n events. The remaining lines will each consist of n integers with each line representing a student's chronological ordering of the n events. All
lines will contain n numbers in the range
, with each number appearing exactly once per line, and with each number separated
from other numbers on the same line by one or more spaces.
代码
#include <stdio.h>
#include<cstring>
#include<algorithm>
int main()
{
int n;
//freopen("input.txt","r",stdin);
while(~scanf("%d",&n))
{
int a[n+1],b[n+1],dp[n+2][n+2];
for(int i=1;i<=n;i++)scanf("%d",a+i);
while(~scanf("%d",b+1))
{
memset(dp,0,sizeof(dp));
for(int j=2;j<=n;j++)scanf("%d",b+j);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
if(a[i]==b[j])dp[i+1][j+1]=dp[i][j]+1;
else
{
dp[i+1][j+1]=std::max(dp[i][j+1],dp[i+1][j]);
}
}
printf("%d\n",dp[n+1][n+1]);
}
}
return 0;
}
A - History Grading
Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
Submit Status Practice UVA
111
Appoint description:
System Crawler (2014-03-27)
Description
History Grading |
Background
Many problems in Computer Science involve maximizing some measure according to constraints.Consider a history exam in which students are asked to put several historical events into chronological order. Students who order all the events correctly will receive full credit, but how should partial credit
be awarded to students who incorrectly rank one or more of the historical events?
Some possibilities for partial credit include:
1 point for each event whose rank matches its correct rank
1 point for each event in the longest (not necessarily contiguous) sequence of events which are in the correct order relative to each other.
For example, if four events are correctly ordered 1 2 3 4 then the order 1 3 2 4 would receive a score of 2 using the first method (events 1 and 4 are correctly ranked) and a score of 3 using the second method
(event sequences 1 2 4 and 1 3 4 are both in the correct order relative to each other).
In this problem you are asked to write a program to score such questions using the second method.
The Problem
Given the correct chronological order of n eventsas
where
denotes
the ranking of event i in the correct chronological order and a sequence of student responses
where
denotes
the chronological rank given by the student to event i; determine the length of the longest (not necessarily contiguous) sequence of events in the student responses that are in the correct chronological order relative to each other.
The Input
The first line of the input will consist of one integer n indicating the number of events with.
The second line will contain nintegers, indicating the correct chronological order of n events. The remaining lines will each consist of n integers with each line representing a student's chronological ordering of the n events. All
lines will contain n numbers in the range
, with each number appearing exactly once per line, and with each number separated
from other numbers on the same line by one or more spaces.
The Output
For each student ranking of events your program should print the score for that ranking. There should be one line of output for each student ranking.Sample Input 1
4 4 2 3 1 1 3 2 4 3 2 1 4 2 3 4 1
Sample Output 1
1 2 3
Sample Input 2
10 3 1 2 4 9 5 10 6 8 7 1 2 3 4 5 6 7 8 9 10 4 7 2 3 10 6 9 1 5 8 3 1 2 4 9 5 10 6 8 7 2 10 1 3 8 4 9 5 7 6
Sample Output 2
6 5 10 9
代码
#include <stdio.h>
#include<cstring>
#include<algorithm>
int main()
{
int n;
//freopen("input.txt","r",stdin);
while(~scanf("%d",&n))
{
int a[n+1],b[n+1],dp[n+2][n+2];
for(int i=1;i<=n;i++)scanf("%d",a+i);
while(~scanf("%d",b+1))
{
memset(dp,0,sizeof(dp));
for(int j=2;j<=n;j++)scanf("%d",b+j);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
if(a[i]==b[j])dp[i+1][j+1]=dp[i][j]+1;
else
{
dp[i+1][j+1]=std::max(dp[i][j+1],dp[i+1][j]);
}
}
printf("%d\n",dp[n+1][n+1]);
}
}
return 0;
}
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