pat 秋 Professional Ability Test

7-4

Professional Ability Test

(30分)

Professional Ability Test (PAT) consists of several series of subject tests. Each test is divided into several levels. Level A is a prerequisite (前置要求) of Level B if one must pass Level A with a score no less than SSS in order to be qualified to take Level B. At the mean time, one who passes Level A with a score no less than SSS will receive a voucher（代金券）of DDD yuans (Chinese dollar) for taking Level B.

At the moment, this PAT is only in design and hence people would make up different plans. A plan is NOT consistent if there exists some test T so that T is a prerequisite of itself. Your job is to test each plan and tell if it is a consistent one, and at the mean time, find the easiest way (with minimum total SSS) to obtain the certificate of any subject test. If the easiest way is not unique, find the one that one can win the maximum total value of vouchers.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers NNN (≤1000\le 1000≤1000) and MMM, being the total numbers of tests and prerequisite relations, respectively. Then MMM lines follow, each describes a prerequisite relation in the following format:

T1 T2 S D

where T1 and T2 are the indices (from 0 to N−1N-1N−1) of the two distinct tests; S is the minimum score (in the range (0, 100]) required to pass T1 in order to be qualified to take T2; and D is the value of the voucher (in the range (0, 500]) one can receive if one passes T1 with a score no less than S and plan to take T2. It is guaranteed that at most one pair of S and D are defined for a prerequisite relation.

Then another positive integer KKK (≤N\le N≤N) is given, followed by KKK queries of tests. All the numbers in a line are separated by spaces.

Output Specification:

Print in the first line Okay. if the whole plan is consistent, or Impossible. if not.

If the plan is consistent, for each query of test T, print in a line the easiest way to obtain the certificate of this test, in the format:

T0->T1->...->T

However, if T is the first level of some subject test (with no prerequisite), print You may take test T directly. instead.

If the plan is impossible, for each query of test T, check if one can take it directly or not. If the answer is yes, print in a line You may take test T directly.; or print Error. instead.

Sample Input 1:

8 15
0 1 50 50
1 2 20 20
3 4 90 90
3 7 90 80
4 5 20 20
7 5 10 10
5 6 10 10
0 4 80 60
3 1 50 45
1 4 30 20
1 5 50 20
2 4 10 10
7 2 10 30
2 5 30 20
2 6 40 60
8
0 1 2 3 4 5 6 7

Sample Output 1:

Okay.
You may take test 0 directly.
0->1
0->1->2
You may take test 3 directly.
0->1->2->4
0->1->2->4->5
0->1->2->6
3->7

Sample Input 2:

Sample Output 2:

Impossible.
You may take test 3 directly.
Error.

这题只弄到6分秋季这个题受益颇多第一题熊猫那个还没做这又把晚上搭上了终于过了样例卑微的要求哈哈

这题主要是信息量大有些东西没读清比如用了个最高级让我成功忽略了第一标准还是没练到位

还有没有去标记天真啊

同起点环的判断以及更新最大值忘记更新两个标准只要答案改变相应的另一个参数值也要改变

还有两个标准的存储与前后两个节点有关不是只有一个结点有关想当然采用hash 位数存储效果不错

#include<bits/stdc++.h>
using namespace std;
int book[1000];//crt
vector<int> temp,ans[1000],a[1000];//crt
//zl外面入好第一个 temp  标记
map<int,int> yr,ff;
int maxsum=-1,minsc=1000000000;int cnt=0,flagg=0;
void dfs(int index){//endif(a[index].size()==0){int sum=0,sc=0;for(int i=1;i<temp.size();i++){sum+=yr[temp[i-1]*1000+temp[i]];sc+=ff[temp[i-1]*1000+temp[i]];//debug//printf(" %d",yr[temp[i-1]*1000+temp[i]]); }if(sc<minsc){minsc=sc;//debug//printf(" %d",sc);ans[cnt]=temp;maxsum=sum;//另一个标准只要更新了答案也要更新当前答案 }   else if(sc==minsc&&sum>maxsum){//第一标准没用上   If the easiest way is not unique, find the one that one can win the maximum total value of vouchers.首先是一个最高级 ans[cnt]=temp;maxsum=sum;//取最大值又忘记赋值了你要改变最大值 }temp.pop_back();}for(int i=0;i<a[index].size();i++){if(book[a[index][i]]==0){book[a[index][i]]=1;//自己路上  有向图没有说是自己到自己的环到可以有多条不同路径到一个地方的方式 所以还是要去标记 如果在一条路上自己碰到自己不行 temp.push_back(a[index][i]);//debug//printf(" %d",a[index][i]) ;dfs(a[index][i]);book[a[index][i]]=0;}else{flagg=1;//多层递归未必能反回 不如设置一个标记  //判断有无回到自己的环 }}if(a[index].size()!=0)temp.pop_back();
}
int main(){int n,m,_1,_2,_3,_4;scanf("%d%d",&n,&m);for(int i=0;i<m;i++){scanf("%d%d%d%d",&_1,&_2,&_3,&_4);a[_2].push_back(_1);yr[_2*1000+_1]=_4;ff[_2*1000+_1]=_3;}int k,_5;scanf("%d",&k);int flag=1;
map<int,int> ok;for(int i=0;i<k;i++){scanf("%d",&_5);temp.clear();maxsum=-1; minsc=1000000000;flagg=0;fill(book,book+1000,0);temp.push_back(_5);book[_5]=1; dfs(_5);//多层递归返回值可能没用了  还是全局外的标记靠谱注意crt cnt++;if(flagg==1){flag=-1;ok[_5]=1;}}if(flag==-1){printf("Impossible.\n");for(int i=0;i<cnt;i++){if(ok[ans[i][0]]==1){printf("Error.\n");}else{if(ans[i].size()==1){printf("You may take test %d directly.\n",ans[i][0]);}else {for(int j=0;j<ans[i].size();j++){if(j!=0) printf("->");printf("%d",ans[i][ans[i].size()-1-j]);}printf("\n");}}}}else{printf("Okay.\n");for(int i=0;i<cnt;i++){if(ans[i].size()==1){printf("You may take test %d directly.\n",ans[i][0]);}else {for(int j=0;j<ans[i].size();j++){if(j!=0) printf("->");printf("%d",ans[i][ans[i].size()-1-j]);}printf("\n");}}}return 0;
}