友好城市

对北岸（或南岸）的城市从小到大排序，再求南岸（或北岸）的城市位置的最长不下降序列长度即可。

ps:这里数据较弱，用n*n的做法可以过。

Code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
//Mystery_Sky
//
#define M 1000100
struct node{
    int N, S;
}city[M];
int f[M], n, ans;

inline bool cmp(node a, node b)
{
    return a.N < b.N;
}

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d%d", &city[i].N, &city[i].S), f[i] = 1;
    sort(city+1, city+1+n, cmp);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= i; j++) {
            if(city[j].S < city[i].S) f[i] = max(f[i], f[j]+1);
        }
    }
    for(int i = 1; i <= n; i++) ans = max(ans, f[i]);
    printf("%d\n", ans);
    return 0;
}

标签：node,include,1263,友好城市,int,较弱,9.7,南岸
来源： https://www.cnblogs.com/Benjamin-cpp/p/10994943.html