P3957 [NOIP2017 普及组] 跳房子

合集 - 思维题(99)

1.大雪球2024-01-162.E. Final Countdown2024-02-183.D. Divisible Pairs2024-02-194.F. Chat Screenshots2024-02-195.G. One-Dimensional Puzzle2024-02-206.P8775 [蓝桥杯 2022 省 A] 青蛙过河2024-02-217.Watering an Array2024-02-238.Dora and Search2024-02-249.P1137 旅行计划2024-02-2410.P3067 [USACO12OPEN] Balanced Cow Subsets G2024-02-2611.P4799 [CEOI2015 Day2] 世界冰球锦标赛2024-02-2612.A. Brick Wall2024-02-2813.B. Minimize Inversions2024-02-2814.C. XOR-distance2024-02-2815.A. Moving Chips2024-02-2816.B. Monsters Attack!2024-02-2817.C. Find B2024-02-2818.D. Slimes2024-02-2819.C. Turtle Fingers: Count the Values of k2024-02-2820.D. Turtle Tenacity: Continual Mods2024-02-2821.D. Vlad and Division2024-02-2822.C. LR-remainders2024-02-2823.P5322 [BJOI2019] 排兵布阵2024-03-0324.P2946 [USACO09MAR] Cow Frisbee Team S2024-03-0325.P1156 垃圾陷阱2024-03-0326.P1064 [NOIP2006 提高组] 金明的预算方案2024-03-0427.P3047 [USACO12FEB] Nearby Cows G2024-03-0428.P5020 [NOIP2018 提高组] 货币系统2024-03-04

29.P3957 [NOIP2017 普及组] 跳房子2024-03-06

30.P1550 [USACO08OCT] Watering Hole G2024-03-0631.P1525 [NOIP2010 提高组] 关押罪犯2024-03-0732.P1379 八数码难题2024-03-0733.P6121 [USACO16OPEN] Closing the Farm G2024-03-0834.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G2024-03-0835.P9632 [ICPC2020 Nanjing R] K Co-prime Permutation2024-03-0836.P9825 [ICPC2020 Shanghai R] Fibonacci2024-03-0837.P2034 选择数字2024-03-0938.P7072 [CSP-J2020] 直播获奖2024-03-0939.P1714 切蛋糕2024-03-0940.P4147 玉蟾宫2024-03-1041.P2866 [USACO06NOV] Bad Hair Day S2024-03-1042.P2032 扫描2024-03-1043.P1884 [USACO12FEB] Overplanting S2024-03-1044.D. Lonely Mountain Dungeons2024-03-1245.F. Microcycle2024-03-1246.E. Increasing Subsequences__22024-03-1347.D. Valiant's New Map2024-03-1448.Array Repetition2024-03-1449.F. Rudolf and Imbalance2024-03-1450.D. Exam in MAC2024-03-1551.G. Rudolf and Subway2024-03-1552.C. Ehab and Path-etic MEXs2024-03-1753.A. String Transformation 12024-03-1754.D. Solve The Maze2024-03-1855.B. Array Fix2024-03-1856.B. Equal XOR2024-03-1857.F. Maximum White Subtree2024-03-1958.P1652 圆2024-03-2059.P2181 对角线2024-03-2060.P1325 雷达安装2024-03-2061.P3194 [HNOI2008] 水平可见直线2024-03-2062.B - Make Many Triangles2024-03-2363.[ARC172A] Chocolate2024-03-2564.C. Theofanis' Nightmare2024-03-2765.D. Birthday Gift2024-03-2766.C. Arrow Path2024-03-2767.P1265 公路修建2024-03-2868.P1284 三角形牧场2024-03-2869.Kirill and Mushrooms2024-03-2870.C. Sasha and the Casino2024-03-2871.P1470 [USACO2.3] 最长前缀 Longest Prefix2024-04-0172.P5536 【XR-3】核心城市2024-04-0273.P5836 [USACO19DEC] Milk Visits S2024-04-0274.P4391 [BOI2009] Radio Transmission 无线传输2024-04-0475.P3435 [POI2006] OKR-Periods of Words2024-04-0476.P5960 【模板】差分约束2024-04-0477.P3052 [USACO12MAR] Cows in a Skyscraper G2024-04-0578.E. Long Inversions2024-04-0979.C. Deep Down Below2024-04-0980.C. Rings2024-04-0981.F. Unfair Game2024-04-1082.G. GCD on a grid2024-04-1083.C. Inversion Graph2024-04-1084.Increase Subarray Sums2024-04-1085.D. Cross Coloring2024-04-1286.M. Triangle Construction2024-04-1287.D2. Set To Max (Hard Version)2024-04-1288.C. Array Game2024-04-1289.F. Alex's whims2024-04-1290.Queue Sort2024-04-1291.P1908 逆序对2024-04-1392.P5094 [USACO04OPEN] MooFest G 加强版2024-04-1593.P5149 会议座位2024-04-1594.P10337 [UESTCPC 2024] 操作序列2024-04-1595.P10330 [UESTCPC 2024] 黑白珠串2024-04-1596.P10252 线性变换2024-04-1697.P3901 数列找不同2024-04-1698.P10288 [GESP样题 八级] 区间2024-04-1699.P5677 [GZOI2017] 配对统计2024-04-17

收起

原题链接

题解

二分加动态维护区间最大值
注意设立变量的含义，改变变量值的规则

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll sum[500005]={0};
struct unit
{
    ll x,v;
    bool operator <(const unit &b) const{return b.v>v;}//
}room[500005];

ll n,d,k;
ll check(ll g)
{
    memset(sum,-0x3f,sizeof sum);//初始化，由于分数有负，所以赋无限小为未踏入
    sum[0]=0;//如果一步也不走，得到的分数为零

    priority_queue<unit> q;//维护的是线性改变区间内的最大值
    //不能初始放入00，因为虽然放入其他值的时候确保了距离大于d-g，但是零点没有，这就导致了q中违规出现了小于d-g的值0
    ll it=0;
    for(ll i=1;i<=n;i++)//遍历房子
    {
        while(it<i&&room[i].x-room[it].x>=max(1LL,d-g)) q.push({room[it].x,sum[it++]});//可以从哪个房子跳过来，细节，由于要求跳的长度至少为一，所以当d-g小于1的时候要判断
        //但是由于有it<i和递增输入的限制，这个不会触发错误
        //这里添加要放在删除前面，因为刚添加进去的不一定小于d+g,他只是小于前一个点的d+g
        while(q.size()&&room[i].x-q.top().x>d+g) q.pop();

        if(q.size()) sum[i]=max(sum[i],q.top().v+room[i].v);//经过删除和添加后，堆顶是最大值
        //printf("sum[%d] = %d\n",i,sum[i]);
        if(sum[i]>=k)return 1;
    }
    return 0;
}
int main()
{
    cin>>n>>d>>k;

    for(ll i=1;i<=n;i++)  cin>>room[i].x>>room[i].v;//输入为升序

    room[0].x=0;
    ll l=-1,r=1e9+2;

    while(l<r-1)
    {
        ll mid=(l+r)/2;
        //printf("%d\n",mid);
        if(check(mid)) r=mid;
        else l=mid;
        //puts("\n");
    }

    if(l!=1e9+1) cout<<r<<endl;
    else puts("-1");//只有一次大于k都没有才会等于1e9+1
    return 0;
}