最长上升子序列LIS集合 POJ2533，POJ1631,POJ1887,POJ1609

POJ2533  赤裸裸的求最长上升子序列，复杂度为n^2的模版

#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>

#define ll long long

#define eps 1e-7

#define inf 0xfffffff
//const ll INF = 1ll<<61;

using namespace std;

//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;

//#define IN freopen("c:\\Users\\linzuojun\\desktop\\input.txt", "r", stdin)
//#define OUT freopen("c:\\Users\\linzuojun\\desktop\\output.txt", "w", stdout)

int dp[10000 + 5];
int num[10000 + 5];

void clear() {
memset(dp,0,sizeof(dp));
memset(num,0,sizeof(num));
}

int main() {
int n;
while(cin>>n) {
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
cin>>num[i];
int ans = 1;
for(int i=1;i<=n;i++)
dp[i] = 1;
for(int i=2;i<=n;i++) {
int m = 0;
for(int j=1;j<i;j++) {
if(num[i] > num[j] && dp[j] + 1 > dp[i]) {
dp[i] = dp[j] + 1;
}
if(ans < dp[i])
ans = dp[i];
}
}
cout<<ans<<endl;
}
return EXIT_SUCCESS;
}

POJ1631，同样是赤裸裸的LIS，复杂度为nlogn的模版,n^2铁定超时

#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>

#define ll long long

#define eps 1e-7

#define inf 0xfffffff
//const ll INF = 1ll<<61;

using namespace std;

//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;

//#define IN freopen("c:\\Users\\linzuojun\\desktop\\input.txt", "r", stdin)
//#define OUT freopen("c:\\Users\\linzuojun\\desktop\\output.txt", "w", stdout)

int dp[100000 + 5];
int num[100000 + 5];

void clear() {
memset(dp,0,sizeof(dp));
memset(num,0,sizeof(num));
}

int main() {
int t,n;
cin>>t;
while(t--) {
clear();
cin>>n;
for(int i=1;i<=n;i++)
cin>>num[i];
int ans = 0;
for(int i=1;i<=n;i++) {
int tmp = num[i];
int left = 1;
int right = ans;
while(left <= right) {
int mid = (left + right)/2;
if(dp[mid] < tmp)
left = mid + 1;
else
right = mid -1;
}
dp[left] = tmp;
if(ans < left)
ans++;
}
cout<<ans<<endl;
}
return EXIT_SUCCESS;
}

POJ1887，最长不上升子序列，反过来做即可，注意输出格式，前导有个空格，后导两个案例之间有个空行，而且不会提醒PE 只会说WA

#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>

#define ll long long

#define eps 1e-7

#define inf 0xfffffff
//const ll INF = 1ll<<61;

using namespace std;

//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;

//#define IN freopen("c:\\Users\\linzuojun\\desktop\\input.txt", "r", stdin)
//#define OUT freopen("c:\\Users\\linzuojun\\desktop\\output.txt", "w", stdout)

int dp[100000 + 5];
int num[100000 + 5];

void clear() {
memset(dp,0,sizeof(dp));/*
memset(num,0,sizeof(num));*/
}

int main() {
int a;
int Case = 0;
clear();
while(scanf("%d",&a)) {
clear();
int n = 1;
if(a == -1)break;
num
= a;
while(true) {
scanf("%d",&a);
if(a == -1)break;
num[++n] = a;
}
for(int i=1;i<=n;i++)
dp[i] = 1;
int ans = 1;
for(int i=n-1;i>=1;i--) {
for(int j=i+1;j<=n;j++) {
if(dp[j] + 1 > dp[i] && num[i] > num[j])
dp[i] = dp[j] + 1;
if(dp[i] > ans)
ans = dp[i];
}
}
printf("Test #%d:\n",++Case);
printf("  maximum possible interceptions: %d\n\n",ans);
}
return EXIT_SUCCESS;
}

POJ1609，最长不下降子序列，不过是有两个限制条件，先进行排序满足其中一条，再在这个基础上进行 LIS算法即，可注意不等号要改变，还有题目问的是最多，而序列不是给定的，所以自己要先进行排序 

#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>

#define ll long long

#define eps 1e-7

#define inf 0xfffffff
//const ll INF = 1ll<<61;

using namespace std;

//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;

//#define IN freopen("c:\\Users\\linzuojun\\desktop\\input.txt", "r", stdin)
//#define OUT freopen("c:\\Users\\linzuojun\\desktop\\output.txt", "w", stdout)

int dp[100000 + 5];

struct Node {
int x,y;
}node[100000 + 5];

void clear() {
memset(dp,0,sizeof(dp));
memset(node,0,sizeof(node));
}

bool cmp(Node x,Node y) {
if(x.x == y.x)
return x.y < y.y;
return x.x <y.x;
}

int main() {
int n;
while(scanf("%d",&n) == 1) {
if(n == 0) {
puts("*");
break;
}
clear();
for(int i=1;i<=n;i++)
scanf("%d %d",&node[i].x,&node[i].y);
sort(node+1,node+n,cmp);
for(int i=1;i<=n;i++)
dp[i] = 1;
int ans = 1;
for(int i=2;i<=n;i++) {
for(int j=1;j<i;j++) {
if(dp[j] + 1 > dp[i] && node[i].x >= node[j].x && node[i].y >= node[j].y)
dp[i] = dp[j] + 1;
if(dp[i] > ans)
ans = dp[i];
}
}
printf("%d\n",ans);
}
return EXIT_SUCCESS;
}

还有另一种做法，因为范围还是比较小的，所以可以直接dp做

#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>

#define ll long long

#define eps 1e-7

#define inf 0xfffffff
//const ll INF = 1ll<<61;

using namespace std;

//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;

//#define IN freopen("c:\\Users\\linzuojun\\desktop\\input.txt", "r", stdin)
//#define OUT freopen("c:\\Users\\linzuojun\\desktop\\output.txt", "w", stdout)

int num[112][112],dp[112][112];

void clear() {
memset(num,0,sizeof(num));
memset(dp,0,sizeof(dp));
}

int main() {
int n;
while(scanf("%d",&n),n) {
clear();
int x,y;
for(int i=1;i<=n;i++) {
scanf("%d %d",&x,&y);
num[x][y]++;
}
for(int i=1;i<=100;i++)
for(int j=1;j<=100;j++)
dp[i][j] = max(dp[i-1][j],dp[i][j-1]) + num[i][j];
printf("%d\n",dp[100][100]);
}
puts("*");
return EXIT_SUCCESS;
}