poj 1769 Minimizing maximizer 线段树优化的dp

Minimizing maximizer

Time Limit: 5000MS		Memory Limit: 30000K
Total Submissions: 3047		Accepted: 1187

Description

The company Chris Ltd. is preparing a new sorting hardware called Maximizer. Maximizer has n inputs numbered from 1 to n. Each input represents one integer. Maximizer has one output which represents the maximum value present on Maximizer's inputs. 

Maximizer is implemented as a pipeline of sorters Sorter(i1, j1), ... , Sorter(ik, jk). Each sorter has n inputs and n outputs. Sorter(i, j) sorts values on inputs i, i+1,... , j in non-decreasing order and lets the other inputs pass through unchanged. The n-th output of the last sorter is the output of the Maximizer. 

An intern (a former ACM contestant) observed that some sorters could be excluded from the pipeline and Maximizer would still produce the correct result. What is the length of the shortest subsequence of the given sequence of sorters in the pipeline still producing correct results for all possible combinations of input values? 

Task 
Write a program that: 

reads a description of a Maximizer, i.e. the initial sequence of sorters in the pipeline, 
computes the length of the shortest subsequence of the initial sequence of sorters still producing correct results for all possible input data, 
writes the result. 

Input

The first line of the input contains two integers n and m (2 <= n <= 50000, 1 <= m <= 500000) separated by a single space. Integer n is the number of inputs and integer m is the number of sorters in the pipeline. The initial sequence of sorters is described in the next m lines. The k-th of these lines contains the parameters of the k-th sorter: two integers ik and jk (1 <= ik < jk <= n) separated by a single space.

Output

The output consists of only one line containing an integer equal to the length of the shortest subsequence of the initial sequence of sorters still producing correct results for all possible data.

Sample Input

40 6
20 30
1 10
10 20
20 30
15 25
30 40

Sample Output

4

Hint

Huge input data, scanf is recommended.

--------

-------

/** head-file **/

#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>

/** define-for **/

#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)

/** define-useful **/

#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
#define pb push_back
#define mp make_pair

/** test **/

#define Display(A, n, m) {                      \
    REP(i, n){                                  \
        REP(j, m) cout << A[i][j] << " ";       \
        cout << endl;                           \
    }                                           \
}

#define Display_1(A, n, m) {                    \
    REP_1(i, n){                                \
        REP_1(j, m) cout << A[i][j] << " ";     \
        cout << endl;                           \
    }                                           \
}

using namespace std;

/** typedef **/

typedef long long LL;

/** Add - On **/

const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };

const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;

const int MAXN=255111;
struct SegmentTree{
    int num[MAXN];
    struct Tree{
        int l;
        int r;
        int min;
    };
    Tree tree[MAXN*4];
    void push_up(int root){
        tree[root].min=min(tree[root<<1].min,tree[root<<1|1].min);
    }
    void build(int root,int l,int r){
        tree[root].l=l;
        tree[root].r=r;
        if(tree[root].l==tree[root].r){
            tree[root].min=num[l];
            return;
        }
        int mid=(l+r)/2;
        build(root<<1,l,mid);
        build(root<<1|1,mid+1,r);
        push_up(root);
    }
    void update(int root,int pos,int val){
        if(tree[root].l==tree[root].r){
            tree[root].min=val;
            return;
        }
        int mid=(tree[root].l+tree[root].r)/2;
        if(pos<=mid) update(root<<1,pos,val);
        else update(root<<1|1,pos,val);
        push_up(root);
    }
    int query(int root,int L,int R){
        if(L<=tree[root].l&&R>=tree[root].r) return tree[root].min;
        int mid=(tree[root].l+tree[root].r)/2,ret=INF;
        if(L<=mid) ret=min(ret,query(root<<1,L,R));
        if(R>mid) ret=min(ret,query(root<<1|1,L,R));
        return ret;
    }
    void init(int n,int d){
        for (int i=1;i<=n;i++) num[i]=d;
        build(1,1,n);
    }
}D;
const int maxn=255000;
const int maxm=750000;
int n,m;
int S[maxn];
int T[maxm];
int f[maxm];
/**
    f[0][1]=0
    f[0][j]=INF (j>1)
    f[i][j]=f[i-1][j] (j!=T(i-1))
           =min(f[i-1][j],min{f[i][j']|Si<=j'<=Ti}+1) (j==T(i-1))
    f[1]=0
    f[j]=INF (j>1)
    f[Ti]=min(f[Ti],min{Si<=j'<=Ti}+1)
**/
int main()
{
    while (~scanf("%d%d",&n,&m))
    {
        for (int i=1;i<=m;i++) scanf("%d%d",&S[i],&T[i]);
        for (int i=0;i<=n;i++) f[i]=INF;
        D.init(n,INF);
        f[1]=0;
        D.update(1,1,0);
        for (int i=1;i<=m;i++){
            int v=min(f[T[i]],D.query(1,S[i],T[i])+1);
            f[T[i]]=v;
            D.update(1,T[i],v);
        }
        printf("%d\n",f[n]);
    }
    return 0;
}