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  • poj 2481 Cows——树状数组&&python

    题意:有$n$头牛,每只牛都有一个范围值$[S, E]$,如果对于牛i和牛j来说,它们的测验值满足下面的条件则证明牛i比牛j强壮:$S_i leq S_j and E_j leq E_i   and E_i-S_i > E_j-S_j$。请输出每头牛有几头牛比其强壮。

    分析:

    我们可以对$S$按从小到大排序,$E$按从大到小排序

    如果当前所求的$i$的$s[i]==s[i-1] and e[i]==e[i-1]$,那么$ans[i]=ans[i-1]$,

    否则便是查询排序后在他前面的牛的$e$大于等于他的数目(可以利用树状数组去维护)

    #include<algorithm>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    
    const int maxn = 100000 + 10;
    
    struct Node{
        int s, e, id;
    }cows[maxn];
    int C[maxn], num, ans[maxn], n;
    
    bool cmp(Node a, Node b)
    {
        if(a.s == b.s)  return a.e > b.e;
        else  return a.s < b.s;
    }
    
    int lowbit(int x)
    {
        return x & -x;
    }
    int sum(int x)
    {
        int ret = 0;
        while (x > 0)
        {
            ret += C[x];
            x -= lowbit(x);
        }
        return ret;
    }
    void add(int x, int d)
    {
        while (x <= num)
        {
            C[x] += d;
            x += lowbit(x);
        }
    }
    
    int main()
    {
        while(scanf("%d", &n) == 1 && n)
        {
            num = 0;
            for(int i = 0;i < n;i++)
            {
                scanf("%d%d", &cows[i].s, &cows[i].e);
                cows[i].id = i;
                num = max(num, cows[i].e);
            }
            sort(cows, cows+n, cmp);
            memset(C,0,sizeof(C));
            for(int i = 0;i < n;i++)
            {
                if(i == 0)  ans[cows[i].id] = 0;
                if(cows[i].s == cows[i-1].s && cows[i].e == cows[i-1].e)
                    ans[cows[i].id] = ans[cows[i-1].id];
                else
                    ans[cows[i].id] = i - sum(cows[i].e-1);
                add(cows[i].e, 1);
            }
            for(int i = 0;i < n;i++)
                printf("%d%c", ans[i], i == n-1 ? '
    ' : ' ');
        }
        return 0;
    }
    /*
    3
    1 1
    1 2
    2 3
    */

    python,结构体排序用list模拟

    import functools
    
    maxn = 100000 + 10;
    C = [0]*maxn
    num = 0
    ans = [0]*maxn
    n = 0
    
    def cmp(a, b):
        if a[0] == b[0]:  
            if a[1] > b[1]:
                return -1
            elif a[1] < b[1]:
                return 1
            else:
                return 0
        else:  
            if a[0] < b[0]:
                return -1
            else:
                return 1
    
    def lowbit(x):
        return x & -x
    
    def sum(x):
        ret = 0
        while x > 0:
            ret += C[x]
            x -= lowbit(x)
        return ret
    
    def add(x, d):
        while x <= num:
            C[x] += d
            x += lowbit(x)
    
    while(True):    
        n = int(input().strip())
        if n == 0:
            break
        num = 0
        cows = [[0 for col in range(3)] for row in range(n)]
        for i in range(n):
            cows[i][0], cows[i][1] = map(int, input().strip().split())
            cows[i][2] = i
            num = max(num, cows[i][1])
        
        cows = sorted(cows, key=functools.cmp_to_key(cmp))
        #print(cows)
        for i in range(0, num+1):  C[i] = 0
        for i in range(n):
            if i == 0:  ans[cows[i][2]] = 0
            if cows[i][0] == cows[i-1][0] and cows[i][1] == cows[i-1][1]:
                ans[cows[i][2]] = ans[cows[i-1][2]]
            else:
                ans[cows[i][2]] = i - sum(cows[i][1]-1)
            add(cows[i][1], 1)
        
        for i in range(n):
            print(ans[i])
    View Code
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  • 原文地址:https://www.cnblogs.com/lfri/p/13264845.html
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