UPC 5130 Concerts
标签：动态规划
Description
John enjoys listening to several bands, which we shall denote using A through Z. He wants to attend several concerts, so he sets out to learn their schedule for the upcoming season.
He finds that in each of the following n days (1 ≤ n ≤ \(10^5\)), there is exactly one concert. He decides to show up at exactly k concerts (1 ≤ k ≤ 300), in a given order, and he may decide to attend more than one concert of the same band.
However, some bands give more expensive concerts than others, so, after attending a concert given by band b, where b spans the letters A to Z, John decides to stay at home for at least hb days before attending any other concert.
Help John figure out how many ways are there in which he can schedule his attendance, in the desired order. Since this number can be very large, the result will be given modulo \(10^9 + 7.\)
Input
The first line contains k and n. The second line contains the 26 hb values, separated by spaces.
The third line contains the sequence of k bands whose concerts John wants to attend e.g.,AFJAZ, meaning A, then F etc. The fourth line contains the schedule for the following n days,specified in an identical manner.
Output
The number of ways in which he can schedule his attendance (mod 109 + 7).
Sample Input
2 10
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
AB
ABBBBABBBB
Sample Output
10
分析
- 因为每次参加音乐后之后都要等待几天才能参加其他的，所以倒着DP会更方便一些
- 用\(dp[i][j]\)来表示第\(i-n\)场音乐会中，参加了第\(j-k\)个乐队的方案数。
- 可以得到转移方程为，\(dp[i][j]=dp[i+1][j]+dp[i+1+hb(i)][j+1]\),其中\(hb(i)\)为第i场音乐会对应的乐队的hb值。
代码
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#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll MOD=1e9+7;
const int maxn=100005;
int dp[maxn][305];
int w[30];
char s1[maxn],s2[305];
int main(int argc, char const *argv[])
{
int k,n;
scanf("%d%d", &k,&n);
for (int i = 0; i < 26; ++i)
{
scanf("%d", w+i);
}
scanf("%s", s2+1);
scanf("%s", s1+1);
ll ans=0;
for (int i = n; i ; --i)
{
for (int j = k; j ; --j)
{
int &u=dp[i][j];
u=dp[i+1][j];
if(j==k&&s1[i]==s2[j]) u=(u+1)%MOD;
else if(i+w[s1[i]-'A']+1<=n&&s1[i]==s2[j])
u=((ll)dp[i+w[s1[i]-'A']+1][j+1]+(ll)u)%MOD;
}
}
printf("%d\n", dp[1][1]);
return 0;
}