(hdu 6078)2017 Multi-University Training Contest
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题意:
给出一个有n(<=2000)个数字的序列 a(ai <=2000) 再给出一个有m(m<=2000)个数字的序列 b (bi<=2000) ,定义波浪序列为:x1< x2 > x3 < x4(注意第一次必须是上升,不能是下降,也就是说第一项必须是波谷)。现在要求找到一个严格单调递增的序列 f:f1,f2,……fk。以及相对应的严格单调递增的序列g:g1,g2,……gk。(k>=1)使得每个a_fi = b_gi,同时满足a_f1,a_f2,a_f3……a_fk为波浪序列。求不同的fg映射有多少种选取方式。
a,b中分别从前向后选取k个数字。然后相对应的 a 中选择的每个位置的数字要和 b 中选择的对应位次的数字相同。(当然如果a数组出现过x,而b没有出现过x,显然x不可能被选取),而 f 、g 则是相对应的下标。要满足选取出来的这个数字序列是一个波浪序列。显然波浪序列中的数字分成两种:波峰和波谷。
总体来说,这个题就是a、b数组之间的匹配问题,同时满足是一个波浪序列
题解:
用波峰和波谷做为转移状态的手段,具体的看详解
里面写的很详细
代码:
#include <algorithm>#include <bitset>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <iostream>#include <map>#include <queue>#include <set>#include <stack>#include <string>using namespace std;#define clr(a, b) memset(a, b, sizeof(a))#define pb(a) push_back(a)#define fir first#define se second#define LL long long#define ll long longtypedef pair<int, int> pii;typedef pair<LL, int> pli;typedef pair<LL, LL> pll;const LL inf = 0x3f3f3f3f3f3f3f3f;const int maxn = 300100;double eps = 0.000001;double PI = acos(-1);LL mod = 998244353;;LL sum[2010][2],dp[2010][2];int a[2010],b[2010],n,m;int main() { int t; scanf("%d",&t); while(t--) { int n,m; scanf("%d%d",&n,&m); for(int i = 1;i <= n;i++) scanf("%d",&a[i]); for(int i = 1;i <= m;i++) scanf("%d",&b[i]); clr(sum,0); clr(dp,0); LL ans = 0; for(int i = 1;i <= n;i++) { LL cnt1 = 1,cnt0 = 0; for(int j = 1;j <= m;j++) { dp[j][0] = dp[j][1] = 0; if(a[i] == b[j]) { dp[j][0] = cnt1; dp[j][1] = cnt0; ans = (ans + cnt1 + cnt0) % mod; } else if(b[j] < a[i]) cnt0 = (cnt0+sum[j][0])%mod; else cnt1 = (cnt1+sum[j][1])%mod; } for(int j = 1;j <= m;j++) { if(a[i] == b[j]) { sum[j][0] = (sum[j][0] + dp[j][0]) % mod; sum[j][1] = (sum[j][1] + dp[j][1]) % mod; } } } printf("%I64d\n",ans); }}
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