uva 11766 Racing Car Computer(DP)

F	Racing Car Computer Input: Standard Input Output: Standard Output

The racing cars of today are equipped with so many sophisticated equipment. Introduction of a new visual transducer which is interfaced with the on-board computer can tell you on the fly how many cars are ahead of you while how many are trailing. There are
N cars on a racing track. Each has an on-board computer with the new feature. During the race, every single car's computer keeps displaying two integers,
a (The number of cars in front of him) & b (The number of cars behind him) for a particular moment. It is possible that at some time, some of the cars are racing side by side i.e. they are exactly at the same location. A car
will not consider any other car at the same location to be a leading or trailing car.

Now, it is suspected that some of the new transducers are not working properly inside such high speed vehicles. The report with all computers' data generated at a particular timestamp is reported to you. You are to determine the minimum number of cars that
have faulty data.

Input
Each test case begins with an integer N (1<=N<=1000), the number of cars on a track. The next N lines each has two integers -
a & b (0<=a,b<=1500) for a particular car.

The last test case is followed by a line with a single 0 indicating the end of input.

Output

For each test case, print a line in the format, “Case X: Y”, where
X is the case number & Y is the minimum number of cars that must have faulty data according to the report.

Sample Input Output for Sample Input

4 2 2 0 0 0 2 3 1 1 1 1 0

Case 1: 3 Case 2: 1

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 2010;
int dp[maxn][maxn];
int vis[maxn][maxn] = {0}, v , s;
struct car{
int p1 , p2;
car(int a = 0, int b = 0){
p1 = a , p2 = b;
}
}C[maxn];
int N;

bool cmp(car c1 , car c2){
if(c1.p2 == c2.p2){
return c1.p1<c2.p1;
}
return c1.p2<c2.p2;
}

void initial(){
s = 0;
}

void readcase(){
int f , b , cur = 1;
for(int i = 1;i <= N;i++){
scanf("%d%d" , &f , &b);
if(f+b >= N) s++;
else C[cur++] = car(f+1 , N-b);
}
sort(C+1 , C+cur , cmp);
}

int DP(int k ,int i , int sum){
if(k > N-s) return 0;
if(vis[k][i] == v) return dp[k][i];
vis[k][i] = v;
int tem = 1e9;
if(C[i].p1 == C[k].p1 && C[i].p2 == C[k].p2 && sum < C[i].p2-C[i].p1+1){
dp[k][i] = DP(k+1 , i , sum+1);
return dp[k][i];
}
if(C[i].p2 < C[k].p1){
tem = min(DP(k+1 , i , sum)+1 , DP(k+1 , k , 1));
}else{
tem = min(tem , DP(k+1 , i , sum)+1);
}
return dp[k][i] = tem;
}

int main(){
int c = 1;
v = 1;
while(cin >> N && N){
initial();
readcase();
printf("Case %d: %d\n" , c++ , s+DP(1 ,0 , 1));
v++;
}
return 0;
}