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  • 备战NOIP——模板复习17

    这里只有模板,并不作讲解,仅为路过的各位做一个参考以及用做自己复习的资料,转载注明出处。

    逆序数

    树状数组

    /*Copyright: Copyright (c) 2018
    *Created on 2018-11-05 
    *Author: 十甫
    *Version 1.0 
    *Title: 树状数组求逆序对
    *Time: inf mins
    */
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    const int size = 500005;
    typedef long long ll;
    
    struct num {
    	int i, k;
    	bool operator < (const num cmp) const {
    		return k < cmp.k;
    	}
    } data[size];
    int n;
    int bit[size], ord[size];
    inline int lowbit(int x) {
    	return x & (-x);
    }
    void add(int a[], int pos, int k) {
    	while(pos <= n) {
    		a[pos] += k;
    		pos += lowbit(pos);
    	}
    }
    inline int query(int a[], int pos) {
    	int res = 0;
    	while(pos) {
    		res += a[pos];
    		pos -= lowbit(pos);
    	}
    	return res;
    }
    
    int main() {
    	while(scanf("%d", &n) && n) {
    		memset(bit, 0, sizeof(bit));
    		memset(ord, 0, sizeof(ord));
    		for(int i = 1;i <= n;i++) {
    			int k;
    			scanf("%d", &k);
    			data[i] = (num) {i, k};
    		}
    		sort(data + 1, data + 1 + n);
    		int cnt = 1;
    		for(int i = 1;i <= n;i++) {
    			if(i != 1 && data[i].k != data[i - 1].k) {
    				cnt++;
    			}
    			ord[data[i].i] = cnt;
    		}
    		ll ans = 0;
    		for(int i = 1;i <= n;i++) {
    			add(bit, ord[i], 1);
    			ans += (i - query(bit, ord[i]));
    		}
    		printf("%lld
    ", ans);
    	}
    	return 0;
    }

    归并排序

    /*Copyright: Copyright (c) 2018
    *Created on 2018-11-05
    *Author:  Kaizyn_洛谷题解
    *Version 1.0 
    *Title: 归并排序求逆序对
    *Time: inf mins
    */
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    
    using namespace std;
    
    const int Maxn = 5e5 + 7;
    
    int n;
    long long ans;
    int a[Maxn], tmp[Maxn];
    
    void merge_sort(int l, int r) // [L, R)
    {
        if(r - l <= 1) return;
        int mid = (l + r) >> 1;
        merge_sort(l,mid); // [L, Mid)
        merge_sort(mid,r); // [Mid, R)
    
        int i = l, j = mid, pos = l;
        while(pos < r)
        {
            if(i >= mid || j < r && a[j] < a[i] )
            {
                tmp[pos++] = a[j++];
                ans += mid - i;
            }
            else tmp[pos++] = a[i++];
        }
        memcpy(a+l,tmp+l,sizeof(int)*(r-l));
        // for(int k = l; k < r; ++k) a[k] = tmp[k];
    }
    
    int main()
    {
        while( scanf("%d",&n) && n !=0 )
        {
            for(register int i = 0; i < n; ++ i) scanf("%d",a+i);
            ans = 0;
            merge_sort(0,n);
            printf("%lld
    ",ans);
        }
        return 0;
    }
    NOIP 2018 RP++
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  • 原文地址:https://www.cnblogs.com/Black-S/p/9930706.html
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