SOJ2505: The County Fair题解动态规划

2505: The County Fair

Submit your solution     Discuss this problem     Best solutions

Description

Every year, Farmer John loves to attend the county fair. The fair has N booths (1 <= N <= 400), and each booth i is planning to give away a fabulous prize at a particular time P(i) (0 <= P(i) <= 1,000,000,000) during the day. Farmer John has heard about this and would like to collect as many fabulous prizes as possible to share with the cows. He would like to show up at a maximum possible number of booths at the exact times the prizes are going to be awarded.

Farmer John investigated and has determined the time T(i,j) (always in range 1..1,000,000) that it takes him to walk from booth i to booth j. The county fair's unusual layout means that perhaps FJ could travel from booth i to booth j by a faster route if he were to visit intermediate booths along the way. Being a poor map reader, Farmer John never considers taking such routes -- he will only walk from booth i to booth j in the event that he can actually collect a fabulous prize at booth j, and he never visits intermediate booths along the way. Furthermore, T(i,j) might not have the same value as T(j,i) owing to FJ's slow walking up hills.

Farmer John starts at booth #1 at time 0. Help him collect as many fabulous prizes as possible.

Input

The input file contains multiple test cases, for each test case:
* Line 1: A single integer, N.

* Lines 2..1+N: Line i+1 contains a single integer, P(i).

* Lines 2+N..1+N+N^2: These N^2 lines each contain a single integer T(i,j) for each pair (i,j) of booths. The first N of these lines respectively contain T(1,1), T(1,2), ..., T(1,N). The next N lines contain T(2,1), T(2,2), ..., T(2,N), and so on. Each T(i,j) value is in the range 1...1,000,000 except for the diagonals T(1,1), T(2,2), ..., T(N,N), which have the value zero.
Process till EOF.

Output

For each test case, output:
* Line 1: A single integer, containing the maximum number of prizes Farmer John can acquire.

Sample Input

4
13
9
19
3
0
10
20
3
4
0
11
2
1
15
0
12
5
5
13
0

Sample Output

3

Source

U S OPEN, 2006 USACO Contest

 

状态：

d[i]表示到第i个最大值

 

状态转移方程

d[i]=max(d[j]+1)

if(d[j]&&p[i].n>=a[p[j].d][p[i].d]+p[j].n)

 

边界：

if(a[1][p[i].d]<=p[i].n)
   d[i]=1;

代码：

#include<cstdio> #include<algorithm> using namespace std; #define max(a,b) ((a)>(b)?(a):(b)) struct node {int d,n;}p[405]; bool cmp(node a,node b) {return a.n<b.n;} int main() { int n; while(scanf("%d",&n)==1){ int i,j,ans=0,a[405][405],d[405]={0}; for(i=1;i<=n;i++) { scanf("%d",&p[i].n); p[i].d=i; } sort(p+1,p+n+1,cmp); for(i=1;i<=n;i++) for(j=1;j<=n;j++) scanf("%d",a[i]+j); for(i=1;i<=n;i++) if(a[1][p[i].d]<=p[i].n)//i能到达 d[i]=1; for(i=1;i<=n;i++) { for(j=1;j<i;j++) if(d[j]&&d[j]+1>=d[i]&&p[i].n>=a[p[j].d][p[i].d]+p[j].n)//先到j在到i的最大值 d[i]=d[j]+1; ans=max(ans,d[i]); } printf("%d/n",ans); } }