HDU 1025 Constructing Roads In JGShining's Kingdom

动态规划问题，求最长非递减子序列

先用了O(n^2)的算法，结果如预料的一般WA

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define SIZE 500000+5
int a[SIZE];
int res[SIZE];

int main()
{

    freopen("C:\Users\super\Documents\CB_codes\in.txt", "r", stdin);
    //freopen("C:\Users\super\Documents\CB_codes\out.txt","w",stdout);
    int n, m, t, len;
    int fir = 0;
    while( ~ scanf("%d", &n) ) {
        memset(a, 0, sizeof(a) );
        for(int i = 1; i <= n; ++ i) {
            scanf("%d%d", &m, &t);
            a[m] = t;
        }
        len = 1;
        res[1] = 1;
        for(int j = 1; j <= n; ++ j) {
            for(int i = 1; i < j; ++ i) {
                if(a[j] > a[j-1] && res[j] < res[i]+1)
                    res[j] = res[i] + 1;
                if(len < res[j])
                    len = res[j];
            }
        }
        if(len == 1) {
            printf("Case %d:
My king, at most %d road can be built.
", fir+1, len);
        }
        else {
            printf("
Case %d:
My king, at most %d roads can be built.
", fir+1, len);

        }
        fir ++;
    }

    fclose(stdin);
    return 0;
}

之后又查询了O(n*logn)的算法，其中用到了二分查找

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define SIZE 500000+5
int a[SIZE];
int res[SIZE];
int fnd(int a, int l, int h) {
    int M = (l + h) / 2;
    while(l <= h) {
        if(a == res[M])
            return M;
        else {
            if(a > res[M])
                l = M + 1;
            else
                h = M - 1;
            M = (l + h) / 2;
        }
    }
    return l;

}
int main()
{

    freopen("C:\Users\super\Documents\CB_codes\in.txt", "r", stdin);
    //freopen("C:\Users\super\Documents\CB_codes\out.txt","w",stdout);
    int n, m, t, len;
    int fir = 0;
    while( ~ scanf("%d", &n) ) {
        memset(a, 0, sizeof(a) );
        for(int i = 1; i <= n; ++ i) {
            scanf("%d%d", &m, &t);
            a[m] = t;
        }
        len = 1;
        res[1] = a[1];
        for(int j = 2; j <= n; ++ j) {
            t = fnd(a[j], 1, len);
            res[t] = a[j];
            if(t > len)
                len ++;
        }
        if(len == 1) {
            printf("Case %d:
My king, at most %d road can be built.

", fir+1, len);
        }
        else {
            printf("Case %d:
My king, at most %d roads can be built.

", fir+1, len);

        }
        fir ++;
    }

    fclose(stdin);
    return 0;
}