HDU 4417 Super Mario（线段树）

最新推荐文章于 2020-01-14 11:23:49 发布

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最新推荐文章于 2020-01-14 11:23:49 发布

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文章标签： java

原文链接：http://www.cnblogs.com/dacc123/p/8228577.html

本文详细解析了SuperMario游戏中的算法挑战，通过构建一条由砖块组成的路径来模拟马里奥救公主的过程，并使用可持久化线段树解决动态区间查询问题。文章提供了一个完整的C++实现案例。

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Super Mario

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5370 Accepted Submission(s): 2461

Problem Description

Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.

Input

The first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)

Output

For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.

Sample Input

1
10
5 2 7 5 4 3 8 7 7 
8 6
5 0
3 1
9 4
1 0
5 5
5 1
6 3
5 7
7 3

Sample Output

可持久化线段树，求动态区间比某个值小的有几个。时间限制1秒，所以要离散化，否则会超时也会超内存

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <map>

using namespace std;
typedef long long int LL;
const int maxn=1e5;
int l[maxn*18+5];
int r[maxn*18+5];
int rt[maxn+5];
int num[maxn*18+5];
int p;
int n,m;
int ll,rr;
LL H;
LL a[maxn+5];
LL b[maxn+5];
map<LL,int> mm;
int nownode()
{
    l[p]=r[p]=0;
    num[p]=0;
    return p++;
}
void update(int &node,LL begin,LL end,LL tag)
{
    if(!node) node=nownode();
    else
    {
        num[p]=num[node];l[p]=l[node];
        r[p]=r[node];node=p;
        p++;
    }
    if(begin==end)
    {
         num[node]++;
         return ;
    }
    LL mid=(begin+end)>>1;
    if(tag<=mid) update(l[node],begin,mid,tag);
    else update(r[node],mid+1,end,tag);
    num[node]=num[l[node]]+num[r[node]];
}
int query(int node1,int node2,LL begin,LL end,LL tag)

{
    if(0<=begin&&end<=tag)
    {
        return num[node2]-num[node1];
    }
    LL mid=(begin+end)>>1;
    if(tag<=mid) return query(l[node1],l[node2],begin,mid,tag);
    else return num[l[node2]]-num[l[node1]]+query(r[node1],r[node2],mid+1,end,tag);
}
int bin(LL x)
{
    int l=1;int r=n;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(a[mid]<=x)
            l=mid+1;
        else
            r=mid-1;
    }
    return r;
}
int main()
{
    int t;
    scanf("%d",&t);
    LL len=1e9;
    LL x;
    int cas=0;
    while(t--)
    {
        scanf("%d%d",&n,&m);
        memset(rt,0,sizeof(rt));
        memset(num,0,sizeof(num));
        p=1;
        for(int i=1;i<=n;i++)
        {

            scanf("%lld",&a[i]);
            b[i]=a[i];

        }
        sort(a+1,a+n+1);
        int tot=0;
        mm.clear();
        for(int i=1;i<=n;i++)
            if(!mm[a[i]]) 
                mm[a[i]]=tot++;
        for(int i=1;i<=n;i++)
        {
            update(rt[i]=rt[i-1],0,maxn,mm[b[i]]);
        }
        printf("Case %d:\n",++cas);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%lld",&ll,&rr,&H);
            int x=bin(H);
            if(x==0) {printf("0\n");continue;}
            ll++,rr++;
            printf("%d\n",query(rt[ll-1],rt[rr],0,maxn,mm[a[x]]));
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/dacc123/p/8228577.html