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[最长上升子序列]北大 POJ 1631 Bridging signals


[cpp] 
/* THE PROGRAM IS MADE BY PYY */ 
/*----------------------------------------------------------------------------//
    Copyright (c) 2012 panyanyany All rights reserved.
 
    URL   : http://poj.org/problem?id=1631
    Name  : 1631 Bridging signals
    Classification : 最长上升子序列
 
    Date  : Wednesday, July 11, 2012
    Time Stage : half an hour
 
    Result:
10420085    panyanyany
1631
Accepted    472K    125MS   C++
1557B   2012-07-11 07:29:31
 
Test Data :
 
Review :
 
//----------------------------------------------------------------------------*/ 
 
#include <stdio.h> 
#include <stdlib.h> 
#include <string.h> 
#include <math.h> 
#include <vector> 
 
#include <algorithm> 
#include <iostream> 
#include <queue> 
#include <set> 
#include <string> 
 
using namespace std ; 
 
#define MEM(a, v)        memset (a, v, sizeof (a))    // a for address, v for value 
#define max(x, y)        ((x) > (y) ? (x) : (y)) 
#define min(x, y)        ((x) < (y) ? (x) : (y)) 
 
#define INF     (0x3f3f3f3f) 
#define MAXN    (40005) 
 
#define L(x)    ((x)<<1) 
#define R(x)    (((x)<<1)|1) 
#define M(x, y) (((x)+(y)) >> 1) 
 
#define DB    // 
 
int a[MAXN], order[MAXN]; 
 
int LIS(int a[], int n) 

    int i, u, l, m, len; 
 
    len = 1; 
    MEM(order, INF); // 找上升子序列时初始化 INF,找下降子序列时初始化为0 
    for (i = 0; i < n; ++i) 
    { 
        u = len; 
        l = 1; 
        while (l <= u) 
        { 
            m = (l + u) >> 1; 
            if (order[m] < a[i]) 
                l = m + 1; 
            else 
                u = m - 1; 
        } 
        if (order[l] > a[i]) 
            order[l] = a[i]; 
        if (len < l) 
            len = l; 
//      printf("^ %d ^", l); 
    } 
 
    return len; 

 
int main() 

    int i, n, tc; 
    while (scanf("%d", &tc) != EOF) 
    { 
        while (tc--) 
        { 
            scanf("%d", &n); 
            for (i = 0; i < n; ++i) 
                scanf("%d", a+i); 
            printf("%d\n", LIS(a, n)); 
        } 
    } 
    return 0; 

作者:panyanyany
补充:软件开发 , C++ ,
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