跟着大佬重新入门DP
数列两段的最大字段和
Maximum sum
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 41231 Accepted: 12879 Description
Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:
Your task is to calculate d(A).
Input
The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.Output
Print exactly one line for each test case. The line should contain the integer d(A).
Sample Input
1 10 1 -1 2 2 3 -3 4 -4 5 -5Sample Output
13Hint
In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.
Huge input,scanf is recommended.
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 50000+7;
int t,n,arr[maxn],sum;
int a[maxn],b[maxn];
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(int i = 1; i <= n ; i++ ) {
scanf("%d",&arr[i]);
}
a[1] = arr[1];
for(int i = 2; i <= n; i ++ ) {
if(a[i-1]<0)
a[i]=arr[i];
else
a[i]= a[i-1]+arr[i];
}
for(int i = 2; i <= n; i ++ ) {
a[i] = max(a[i-1],a[i]);
}
/*********************************/
b[n] = arr[n];
for(int i = n-1; i >= 1; i -- ) {
if(b[i+1]<0)
b[i]=arr[i];
else
b[i]= b[i+1]+arr[i];
}
for(int i = n-1; i >= 1; i -- ) {
b[i] = max(b[i+1],b[i]);
}
int ans = -999999999;
for(int i = 2; i <= n; i ++ ) {
ans = max(a[i-1]+b[i],ans);
}
printf("%d\n",ans);
}
return 0;
}
最长公共上升子序列
Common Subsequence
Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 53882 Accepted: 22384 Description
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
Input
The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.
Output
For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.
Sample Input
abcfbc abfcab programming contest abcd mnpSample Output
4 2 0
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1e3+7;
char str[maxn],ttr[maxn];
int dp[maxn][maxn];
int main()
{
while(~scanf("%s %s",str+1,ttr+1))
{
int n = strlen(str+1);
int m = strlen(ttr+1);
memset(dp,0,sizeof(dp));
dp[1][1] = str[1] == ttr[1];
for(int i = 1; i <= n; i ++ ) {
for(int j = 1; j <= m ; j ++ ) {
if(i == 1 && j == 1) {
continue;
}
if(str[i] == ttr[j]) {
dp[i][j] = dp[i-1][j-1]+1;
}
else {
dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
}
}
}
printf("%d\n",dp[n][m]);
}
return 0;
}
最长上升子序列
Longest Ordered Subsequence
Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 53931 Accepted: 24094 Description
A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).
Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.Input
The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000
Output
Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.
Sample Input
7 1 7 3 5 9 4 8Sample Output
4
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1005;
int n,arr[maxn],dp[maxn];
int main()
{
while(~scanf("%d",&n))
{
for(int i =1 ; i <= n ; i ++ ) {
scanf("%d",&arr[i]);
dp[i] = 1;
}
for(int i = 1 ; i <= n ; i ++ ) {
for(int j = 1; j < i ; j ++ ) {
if(arr[i]>arr[j]) {
dp[i] = max(dp[i],dp[j]+1);
}
}
}
int ans = 0;
for(int i = 1; i <= n; i ++ ) {
ans = max(ans,dp[i]);
}
printf("%d\n",ans);
}
return 0;
}
//二分版本
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1005;
int n,arr[maxn],dp[maxn],top,data;
int main()
{
while(~scanf("%d",&n))
{
top = 1;
scanf("%d",&dp[0]);
for(int i =1 ; i < n ; i ++ ) {
scanf("%d",&data);
if(data > dp[top-1]) {
dp[top++] = data;
}
else {
dp[lower_bound(dp,dp+top,data)-dp] = data;
}
}
printf("%d\n",top);
}
return 0;
}
最长公共上升子序列
Greatest Common Increasing Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8181 Accepted Submission(s): 2644
Problem Description
This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.
Input
Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.
Output
output print L - the length of the greatest common increasing subsequence of both sequences.
Sample Input
1 5 1 4 2 5 -12 4 -12 1 2 4Sample Output
2
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int f[maxn];
int a[maxn],n;
int b[maxn],m;
int main()
{
int t;
while(~scanf("%d",&t))
{
while(t--)
{
memset(f,0,sizeof(f));
scanf("%d",&n);
for(int i = 1; i <= n; i ++ )
{
scanf("%d",&a[i]);
}
scanf("%d",&m);
for(int i = 1; i <= m; i ++ )
{
scanf("%d",&b[i]);
}
for(int i = 1; i <= n; i ++ )
{
int MAX = 0;
for(int j = 1; j <= m; j ++ )
{
if(a[i] > b[j]) MAX = max(MAX,f[j]);
if(a[i] == b[j]) f[j] = MAX + 1;
}
}
int ans = 0;
for(int j = 1; j <= m; j ++ )
{
ans = max(ans,f[j]);
}
printf("%d\n",ans);
if(t){
puts("");
}
}
}
return 0;
}
