hdu 1025 Constructing Roads In JGShining's Kingdom (LIS)

小记：这题思考了很久，唉~ 最后还是没摸出LIS来...

思路：求出LIS（最长递增子序列）的长度即可。

不过要注意输出，如果只有一个的话，road 就没有s，否则就是roads

关于LIS的O(nlogn)的算法，请参考：LIS（三大算法）

代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <string.h>
#include <stdlib.h>
#include <map>
#include <set>

using namespace std;

#define REP(a,b,c) for(int a = b; a < c; ++a)

const int MAX_ = 500010;
const int INF = 0x7fffffff;
const int N = 500010;

struct node{
    int id, v;
}p[MAX_], q[MAX_];

int d[MAX_];
int B[MAX_];

bool cmp(const node &a, const node &b){
    if(a.id < b.id)return true;
    else return false;
}

int LIS(int d[], int T){
    int *B = new int[T+1];
    int l, r, mid, len = 0;
    B[0] = d[1];
    REP(i, 2, T+1){
        if(d[i] > B[len]){B[++len] = d[i];continue;}
        l = 0; r = len;
        while(l <= r){
            mid = l + (r-l)/2;
            if(B[mid] < d[i])l = mid + 1;
            else r = mid - 1;
        }
        B[l] = d[i];
        //if(l > len) len ++;
    }
    delete[] B;
    return len + 1;
}

int main()
{
    int T, s, t, Ca = 1;
    while(~scanf("%d",&T)){
        REP(i, 0, T){
            scanf("%d %d", &s,&t);
            p[s].v = t;
            q[t].v = s;
            p[s].id = s;
            q[t].id = t;
        }
        sort(p+1, p+T+1, cmp);
        REP(i, 1, T+1){
            d[i] = p[i].v;
        }
        int cnt;
        cnt = LIS(d, T);
        /*REP(i, 1, T+1){
            printf("%d %d\n",p[i].id, p[i].v);
        }
        printf("\n");*/
        sort(q+1, q+T+1, cmp);
        REP(i, 1, T+1){
            d[i] = q[i].v;
        }
        int ans;
        ans = LIS(d, T);

        if(ans > cnt)cnt= ans;

        printf("Case %d:\n", Ca++);

        if(ans == 1)
            printf("My king, at most %d road can be built.\n\n", cnt);
        else printf("My king, at most %d roads can be built.\n\n", cnt);
    }
    return 0;
}

ans 和cnt的值是一样的