POJ 2452 Sticks Problem (暴力或者rmq+二分)

题意：给你一组数a[n]，求满足a[i] < a[k] < a[j] (i <= k <= j)的最大的 j - i 。

析：在比赛时，我是暴力做的，虽然错了好多次，后来说理解是rmq，我又用rmq写了一次，发现rmq还没有我暴力快，rwq 2000多，暴力才700.

暴力中加了一个优化条件就是前枚举 i 时，下一个 i 值不一定是i+1，而是满足条件中的最大值的位置。这样优化就是时间很短了。

如果用rmq，就得用两个dp数组分别记录最大值和最小值的下标，然后枚举 i，在i+1 - n-1这个区间中求第一个小于 a[i] 的数，然后再从 i+1 - 该数，

求最大的那个数的下标。不断更新答案即可。

代码如下：

暴力的代码：

#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>

#define debug puts("+++++")

//#include <tr1/unordered_map>

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

//using namespace std :: tr1;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const LL LNF = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 5e4 + 5;

const LL mod = 1e9 + 7;

const int N = 1e6 + 5;

const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};

const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};

const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }

inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }

inline int lcm(int a, int b){  return a * b / gcd(a, b); }

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 int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

int a[maxn];

int main(){

    while(scanf("%d", &n) == 1){

        for(int i = 0; i < n; ++i)  scanf("%d", a+i);

        int ans = 0;

        int x;

        for(int i = 0; i < n-ans; i = x+1){

            x = Min(x, a[i]);

            int mmax = a[i];

            x = i;

            for(int j = i+1; j < n; ++j){

                if(mmax < a[j]){

                    x = j;

                    mmax = a[j];

                }

                if(a[j] < a[i]) break;

                if(mmax <= a[j]){ ans = Max(ans, j-i); }

            }

        }

        printf("%d\n", ans ? ans : -1);

    }

    return 0;

}

rmq的代码：

#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>

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

//#include <tr1/unordered_map>

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

//using namespace std :: tr1;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const LL LNF = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 5e4 + 5;

const LL mod = 10000000000007;

const int N = 1e6 + 5;

const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};

const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};

const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }

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 int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

int dp1[maxn][20], dp2[maxn][20];

int a[maxn];

inline int Min_(int i, int j){ return a[i] < a[j] ? i : j; }

inline int Max_(int i, int j){ return a[i] < a[j] ? j : i; }

void rmq_init(){

    for(int i = 0; i < n; ++i)  dp1[i][0] = dp2[i][0] = i;

    for(int j = 1; (1<<j) <= n; ++j)

        for(int i = 0; i + (1<<j) - 1 < n; ++i){

            dp1[i][j] = Min_(dp1[i][j-1], dp1[i+(1<<(j-1))][j-1]);

            dp2[i][j] = Max_(dp2[i][j-1], dp2[i+(1<<(j-1))][j-1]);

        }

}

inline int rmqmin(int l, int r){

    int k = (int)(log(r-l+1.0) / log(2.0));

    return Min_(dp1[l][k], dp1[r-(1<<k)+1][k]);

}

inline int rmqmax(int l, int r){

    int k = (int)(log(r-l+1.0) / log(2.0));

    return Max_(dp2[l][k], dp2[r-(1<<k)+1][k]);

}

int solve(int i){

    int l = i+1, r = n-1;

    while(l < r){

        int mid = (l+r) >> 1;

        if(a[rmqmin(l, mid)] > a[i]) l = mid + 1;

        else r = mid;

    }

    return rmqmax(i, l);

}

int main(){

    while(scanf("%d", &n) == 1){

        for(int i = 0; i < n; ++i) scanf("%d", a+i);

        rmq_init();

        int ans = 0;

        for(int i = 0; i < n - ans - 1; ++i){

             int j = solve(i);

             ans = Max(ans, j-i);

        }

        printf("%d\n", ans ? ans : -1);

    }

    return 0;

}

巴特西

POJ 2452 Sticks Problem (暴力或者rmq+二分)

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