BZOJ2298 [HAOI2011]problem a

先把所有人从小到大排序，则

如果一个人说a个人比他大，b个人比他小，等价于他声称[b + 1, n - a]的值是相等的

问题转化为给一些线段，找出一些不重叠的线段，价值最大 (注意：线段可以重合！！！)

然后就是一遍扫过去的DP

f[i]表示以线段i为结尾的满足条件的线段最多数

f[i] = max(f[j]) + cnt[i], 其中r[j] < l[i], cnt[i]表示i这条线段重合次数

DP要优化的说，不过只要二分一下嘛就好了！

 

/**************************************************************
    Problem: 2298
    User: rausen
    Language: C++
    Result: Accepted
    Time:792 ms
    Memory:5012 kb
****************************************************************/
 
#include <cstdio>
#include <algorithm>
#include <utility>
#include <map>
  
using namespace std;
typedef pair<int, int> P;
const int N = 100005;
struct data{
    int x, y, cnt;
    bool f;
}a[N];
inline bool operator < (const data a, const data b){
    return a.y == b.y ? a.x < b.x : a.y < b.y;
}
map <P, int> M;
int n, f[N], X, Y, l;
  
inline int read(){
    int x = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9'){
        if (ch == '-') sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9'){
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return sgn * x;
}
  
int main(){
    n = read();
    for (int i = 1; i <= n; ++i){
        X = read(), Y = read();
        a[i].f = (X + Y >= n);
        a[i].x = n - X, a[i].y = ++Y;
        if (a[i].x > a[i].y) swap(a[i].x, a[i].y);
        a[i].cnt = min(M[make_pair(a[i].x, a[i].y)] += a[i].f ^ 1, a[i].y - a[i].x + 1);
    }
    sort(a + 1, a + n + 1);
    for (int i = 1; i <= n; ++i){
        if (a[i].f){
            f[i] = f[i - 1];
            continue;
        }
        l = lower_bound(a + 1, a + i, (data){-1, a[i].x}) - a - 1;
        f[i] = max(f[i - 1], f[l] + a[i].cnt);
    }
    printf("%d\n", n - f[n]);
    return 0;
}