BZOJ4723: [POI2017]Flappy Bird

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每个时间点,鸟一定会向上或下飞一格,能飞到的上下界都变化1,且奇偶性一定改变,因此能飞到的地方是上下界内符合奇偶性的高度
扫描线扫过去,可以O(1)算得从last到x上下界的变化,和当前飞过的区间取个交(注意奇偶性),对于最小次数,高度确定,点击次数就确定,取下界

code:

#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long long#define inf 1e9using namespace std;inline void down(int &x,const int &y){if(x>y)x=y;}inline void up(int &x,const int &y){if(x<y)x=y;}const int maxn = 510000;int now,L,R,X,n;int main(){    scanf("%d%d",&n,&X);    now=L=R=0;    bool flag=true;    for(int i=1;i<=n;i++)    {        int x,li,ri; scanf("%d%d%d",&x,&li,&ri);        li++; ri--;        L=L-(x-now);        R=R+(x-now);        up(L,li); down(R,ri);        if(x&1)        {            if(!(L&1)) L++;            if(!(R&1)) R--;        }        else        {            if(L&1) L++;            if(R&1) R--;        }        if(L>R) { flag=false; break; }        now=x;    }    if(flag) printf("%d\n",(L-(-now))>>1);    else puts("NIE");    return 0;}
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