POJ 1036 Gangsters

Gangsters

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 13041 Accepted: 3728

Description

N gangsters are going to a restaurant. The i-th gangster comes at the time Ti and has the prosperity Pi. The door of the restaurant has K+1 states of openness expressed by the integers in the range [0, K]. The state of openness can change by one in one unit of time; i.e. it either opens by one, closes by one or remains the same. At the initial moment of time the door is closed (state 0). The i-th gangster enters the restaurant only if the door is opened specially for him, i.e. when the state of openness coincides with his stoutness Si. If at the moment of time when the gangster comes to the restaurant the state of openness is not equal to his stoutness, then the gangster goes away and never returns.
The restaurant works in the interval of time [0, T].
The goal is to gather the gangsters with the maximal total prosperity in the restaurant by opening and closing the door appropriately.

Input

?The first line of the input file contains the values N, K, and T, separated by spaces. (1 <= N <= 100 ,1 <= K <= 100 ,0 <= T <= 30000 )
?The second line of the input file contains the moments of time when gangsters come to the restaurant T1, T2, …, TN, separated by spaces. ( 0 <= Ti <= T for i = 1, 2, …, N)
?The third line of the input file contains the values of the prosperity of gangsters P1, P2, …, PN, separated by spaces. ( 0 <= Pi <= 300 for i = 1, 2, …, N)
?The forth line of the input file contains the values of the stoutness of gangsters S1, S2, …, SN, separated by spaces. ( 1 <= Si <= K for i = 1, 2, …, N)
All values in the input file are integers.

Output

Print to the output file the single integer ?the maximal sum of prosperity of gangsters in the restaurant. In case when no gangster can enter the restaurant the output should be 0.

Sample Input

4 10 20
10 16 8 16
10 11 15 1
10 7 1 8

Sample Output

26

Thinking: 三天来的第一道题，果然食言了。晚上一点睡，痘痘长一对。言归正传，这道题还是比较常规的，可惜在看openness of the door coincides with the stousness 这点的时候晕了 ，纠结了好久。

dp[i]: 表示第i个人进去后的最大P值
可得地推关系式 dp[i] = max(dp[i], dp[j]＋line[i].p);

#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int N, K, T;const int maxn = 105;struct node{    int t, p, s;    bool operator < (const node& n) const{        if(n.t != t){            return t < n.t;        }else{            return s < n.s;        }    }}line[maxn];int dp[maxn];void solve(){    memset(dp, 0, sizeof(dp));    sort(line+1, line+N+1);    for(int i = 1; i <= N; i++){        for(int j = 0; j < i; j++){            if(dp[j] >= line[j].p){//这里一定要判断第J个人进去了                if(abs(line[i].s - line[j].s)  <= line[i].t - line[j].t){                    dp[i] = max(dp[i], dp[j] + line[i].p);                }            }        }    }    int ans = 0;    for(int i = 1; i <= N; i++)        ans = max(ans, dp[i]);    printf("%d\n", ans);}int main(){    while(scanf("%d%d%d", &N, &K, &T) == 3){        for(int i = 1; i <= N; i++)            scanf("%d", &line[i].t);        for(int i = 1; i <= N; i++)            scanf("%d", &line[i].p);        for(int i = 1; i <= N; i++)            scanf("%d", &line[i].s);        solve();    }       return 0;}