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• # 51Nod 1376 最长递增子序列的数量 (DP+BIT)

题意：略。

析：dp[i] 表示以第 i 个数结尾的LIS的长度和数量，状态方程很好转移，先说长度 dp[i] = max { dp[j] + 1 | a[i] > a[j] && j < i }，然后是数量，dp[i] = sigma(dp[j]) if dp[i] == dp[j] + 1。

如果普通转移时间复杂度很高，达不到要求，由于有个求和的操作，可以考虑用BIT优化，先把每个数离散化，然后对每个数只要求小于它的数，并且长度最长的就好了，数量也是，如果长度一样就进行合并，否则不合并，或者更新长度。

代码如下：

```#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e4 + 10;
const int maxm = 1e6 + 5;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}

void update(P &a, P b){
if(a.fi < b.fi)  a = b;
else if(a.fi == b.fi){
a.se += b.se;
if(a.se >= mod)  a.se -= mod;
}
}

P sum[maxn];

inline int lowbit(int x){ return -x&x; }

while(x <= m){
update(sum[x], c);
x += lowbit(x);
}
}

P query(int x){
P ans(0, 1);
while(x){
update(ans, sum[x]);
x -= lowbit(x);
}
return ans;
}
int a[maxn];
vector<int> v;

int getpos(int x){ return lower_bound(v.begin(), v.end(), x) - v.begin() + 1; }

int main(){
scanf("%d", &n);
v.resize(n);
for(int i = 0; i < n; ++i){
scanf("%d", a+i);
v[i] = a[i];
}
sort(v.begin(), v.end());
v.resize(unique(v.begin(), v.end()) - v.begin());
m = v.sz;
P ans(0, 1);
for(int i = 0; i < n; ++i){
int pos = getpos(a[i]);
P tmp = query(pos - 1);
++tmp.fi;
update(ans, tmp);
}
printf("%d
", ans.se);
return 0;
}
```

分治：

```#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e4 + 10;
const int maxm = 1e6 + 5;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}

void update(P &a, P b){
if(a.fi < b.fi)  a = b;
else if(a.fi == b.fi){
a.se += b.se;
if(a.se >= mod)  a.se -= mod;
}
}
P dp[maxn], res;

int id[maxn], a[maxn];

inline bool cmp(int x, int y){ return a[x] == a[y] ? x > y : a[x] < a[y];                                                                                              }

void dfs(int l, int r){
if(l == r){ update(dp[l], P(1, 1));  return ; }
int m = l + r >> 1;
dfs(l, m);
for(int i = l; i <= r; ++i)  id[i] = i;
sort(id+l, id+r+1, cmp);

P ans = P(0, 0);
for(int i = l; i <= r; ++i){
int idx = id[i];
if(idx <= m)  update(ans, dp[idx]);
else{
P tmp = ans;
++tmp.fi;
update(dp[idx], tmp);
update(res, tmp);
}
}
dfs(m+1, r);
}

int main(){
scanf("%d", &n);
for(int i = 0; i < n; ++i)  scanf("%d", a+i);
dfs(0, n-1);
printf("%d
", res.se);
return 0;
}
```

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