1018. Public Bike Management (30)
There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it to any other stations in the city.
The Public Bike Management Center (PBMC) keeps monitoring the real-time capacity of all the stations. A station is said to be in perfect condition if it is exactly half-full. If a station is full or empty, PBMC will collect or send bikes to adjust the condition of that station to perfect. And more, all the stations on the way will be adjusted as well.
When a problem station is reported, PBMC will always choose the shortest path to reach that station. If there are more than one shortest path, the one that requires the least number of bikes sent from PBMC will be chosen.
Figure 1
Figure 1 illustrates an example. The stations are represented by vertices and the roads correspond to the edges. The number on an edge is the time taken to reach one end station from another. The number written inside a vertex S is the current number of bikes stored at S. Given that the maximum capacity of each station is 10. To solve the problem at S3, we have 2 different shortest paths:
1. PBMC -> S1 -> S3. In this case, 4 bikes must be sent from PBMC, because we can collect 1 bike from S1 and then take 5 bikes to S3, so that both stations will be in perfect conditions.
2. PBMC -> S2 -> S3. This path requires the same time as path 1, but only 3 bikes sent from PBMC and hence is the one that will be chosen.
Input Specification:
Each input file contains one test case. For each case, the first line contains 4 numbers: Cmax (<= 100), always an even number, is the maximum capacity of each station; N (<= 500), the total number of stations; Sp, the index of the problem station (the stations are numbered from 1 to N, and PBMC is represented by the vertex 0); and M, the number of roads. The second line contains N non-negative numbers Ci(i=1,...N) where each Ci is the current number of bikes at Si respectively. Then M lines follow, each contains 3 numbers: Si, Sj, and Tij which describe the time Tij taken to move betwen stations Si and Sj. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print your results in one line. First output the number of bikes that PBMC must send. Then after one space, output the path in the format: 0->S1->...->Sp. Finally after another space, output the number of bikes that we must take back to PBMC after the condition of Sp is adjusted to perfect.
Note that if such a path is not unique, output the one that requires minimum number of bikes that we must take back to PBMC. The judge's data guarantee that such a path is unique.
Sample Input:10 3 3 5 6 7 0 0 1 1 0 2 1 0 3 3 1 3 1 2 3 1Sample Output:
3 0->2->3 0
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int flag[501];
int grond[501][501];
int n,m,aim;
int ma;
int val[501];
int indexans;
vector<int> curoad;
vector<int> miroad;
int bikewehave,minbike;//我們現在有的自行車和答案所求自行車
int milen,curlen,bikewesend,misen;
void dfs(int t){
if(curlen>milen)
return;
if(t == aim){
bool yes = false;
if(curlen<milen){//如果當前路更短
yes = true;
}
else{ //如果當前也是最短路
if(bikewesend<misen)
yes=true;
else if(bikewesend==misen)
if(bikewehave<minbike)
yes=true;
}
if(yes) //如果當前解法更優
{
milen=curlen;
misen=bikewesend;
minbike=bikewehave;
miroad=curoad;
}
}
else
{
for(int i=0;i<=n;i++){
if(!flag[i] && grond[t][i]>0){
flag[i]=1;
curlen+=grond[t][i];
int tebike = bikewehave;
int tesen = bikewesend;
if(val[i]+bikewehave<ma/2){ //如果補不滿當前車站
bikewesend+=ma/2-val[i]-bikewehave; //要送更多車輛
bikewehave=0; //路上撿的車送光啦
}
else
bikewehave+=val[i]-ma/2; //又撿車咯
curoad.push_back(i);
dfs(i);
curoad.pop_back(); //返回dfs前的狀態
bikewehave=tebike;
bikewesend=tesen;
curlen-=grond[t][i];
flag[i]=0;
}
}
}
}
int main()
{
cin>>ma>>n>>aim>>m;
for(int i=1;i<=n;i++){
cin>>val[i];
}
memset(grond,0,sizeof(grond));
for(int i=0;i<m;i++)
{
int a,b,c;
cin>>a>>b>>c;
grond[a][b]=c;
grond[b][a]=c;
}
bikewehave=curlen=bikewesend=0;
minbike=milen=misen=1e9;
memset(flag,0,sizeof(flag));
flag[0]=1;
dfs(0);
printf("%d ",misen);
printf("%d",0);
for(int i=0;i<miroad.size();i++)
{
printf("->%d",miroad[i]);
}
printf(" %d",minbike);
return 0;
}1022. Digital Library (30)
A Digital Library contains millions of books, stored according to their titles, authors, key words of their abstracts, publishers, and published years. Each book is assigned an unique 7-digit number as its ID. Given any query from a reader, you are supposed to output the resulting books, sorted in increasing order of their ID's.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (<=10000) which is the total number of books. Then N blocks follow, each contains the information of a book in 6 lines:
- Line #1: the 7-digit ID number;
- Line #2: the book title -- a string of no more than 80 characters;
- Line #3: the author -- a string of no more than 80 characters;
- Line #4: the key words -- each word is a string of no more than 10 characters without any white space, and the keywords are separated by exactly one space;
- Line #5: the publisher -- a string of no more than 80 characters;
- Line #6: the published year -- a 4-digit number which is in the range [1000, 3000].
It is assumed that each book belongs to one author only, and contains no more than 5 key words; there are no more than 1000 distinct key words in total; and there are no more than 1000 distinct publishers.
After the book information, there is a line containing a positive integer M (<=1000) which is the number of user's search queries. Then M lines follow, each in one of the formats shown below:
- 1: a book title
- 2: name of an author
- 3: a key word
- 4: name of a publisher
- 5: a 4-digit number representing the year
Output Specification:
For each query, first print the original query in a line, then output the resulting book ID's in increasing order, each occupying a line. If no book is found, print "Not Found" instead.
Sample Input:3 1111111 The Testing Book Yue Chen test code debug sort keywords ZUCS Print 2011 3333333 Another Testing Book Yue Chen test code sort keywords ZUCS Print2 2012 2222222 The Testing Book CYLL keywords debug book ZUCS Print2 2011 6 1: The Testing Book 2: Yue Chen 3: keywords 4: ZUCS Print 5: 2011 3: blablablaSample Output:
1: The Testing Book 1111111 2222222 2: Yue Chen 1111111 3333333 3: keywords 1111111 2222222 3333333 4: ZUCS Print 1111111 5: 2011 1111111 2222222 3: blablabla Not Found
C++字符串讀取一行爲getline(cin,stringname);在這之前如果有輸入其他的東西的話需要一個getchar()，不然自動讀入換行符。
#include <string>
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
struct book{
string id;
string title;
string author;
string key;
int kn;
string publisher;
string year;
};
book asd[10086];
int n;
bool cmp(book a,book b){ //排序用
return a.id<b.id;
}
void sov(int i){
string s;
getline(cin,s);
int yes = false;
if(i==1)
{
for(int j=0;j<n;j++){
if(asd[j].title==s){
if(!yes) cout<<i<<": "<<s<<endl;
cout<<asd[j].id<<endl;
yes = true;
}
}
}
else if(i==2){
for(int j=0;j<n;j++){
if(asd[j].author.find(s)!=string::npos){
if(!yes) cout<<i<<": "<<s<<endl;
cout<<asd[j].id<<endl;
yes = true;
}
}
}
else if(i==3){
for(int j=0;j<n;j++){
if(asd[j].key.find(s)!=string::npos){
if(!yes)
cout<<i<<": "<<s<<endl;
cout<<asd[j].id<<endl;
yes = true;
}
}
}
else if(i==4){
for(int j=0;j<n;j++){
if(asd[j].publisher==s){
if(!yes)
cout<<i<<": "<<s<<endl;
cout<<asd[j].id<<endl;
yes = true;
}
}
}
else if(i==5){
for(int j=0;j<n;j++){
if(asd[j].year==s){
if(!yes) cout<<i<<": "<<s<<endl;
cout<<asd[j].id<<endl;
yes = true;
}
}
}
if(!yes)
cout<<i<<": "<<s<<endl<<"Not Found"<<endl;
}
int main(){
cin>>n;getchar();
for(int i=0;i<n;i++){
string s;
getline(cin,s);
asd[i].id=s;
getline(cin,s);
asd[i].title=s;
getline(cin,s);
asd[i].author=s;
getline(cin,s);
asd[i].key=s;
getline(cin,s);
asd[i].publisher=s;
getline(cin,s);
asd[i].year=s;
}
sort(asd,asd+n,cmp);
int m;
cin>>m;
for(int i=0;i<m;i++){
int od;
cin>>od;
getchar();getchar();
sov(od);
}
return 0;
}