HDU 5495 LCS （置换）

题意:

求一个置换,使得置换后的两个序列的LCS最大,求这个最大LCS

分析:

我们手算可以发现第一序列对于第二个序列的置换,可以把序列分成若干个循环节环,显然环与环之间是独立的.
事实上对于一个长度为l(l>1)的环,我们总可以得到一个长度为l−1的LCS
那么答案就是ans=∑max{1,l−1}
记得先排序−−求循环节是从1−n

代码:

////  Created by TaoSama on 2015-10-03//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, a[N], b[N];bool vis[N];int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    while(t--) {        scanf("%d", &n);        for(int i = 1; i <= n; ++i) scanf("%d", a + i);        for(int i = 1; i <= n; ++i) {            int x; scanf("%d", &x);            b[a[i]] = x;        }        memset(vis, false, sizeof vis);        int ans = 0;        for(int i = 1; i <= n; ++i) {            if(!vis[i]) {                int j = i, cnt = 0;                while(!vis[j]) {                    ++cnt;                    vis[j] = true;                    j = b[j];                }                ans += max(1, cnt - 1);            }        }        printf("%d\n", ans);    }    return 0;}