1074 Reversing Linked List （25 point(s)）

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤10​5​​) which is the total number of nodes, and a positive K (≤N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer, and Next is the position of the next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

題目大意：連結串列反轉。

解題思路：其實挺水，有個坑就是反轉的時候要注意。不一定是反轉一次的，可能反轉多次，他的意思不是前K個反轉一次，而是每K個反轉一次。

程式碼如下

#include<iostream>
#include<string.h>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<time.h>
#include<math.h>
#include<set>
#include<list>
#include<climits>
#include<queue>
#include<cstring>
#include<map>
#include<stack>
#include<string>
using namespace std;
struct Node {
	int addr;
	int data;
	int next;
};
map<int, Node> a;
vector<Node> res;
int main() {
	int begin; int N, K;
	scanf("%d %d %d", &begin,&N,&K);
	for (int i = 0; i < N; i++) {
		Node tmp;
		scanf("%d %d %d", &tmp.addr, &tmp.data, &tmp.next);
		a[tmp.addr] = tmp;
	}
	while (true) {
		if (begin == -1)break;
		res.push_back(a[begin]);
		begin = a[begin].next;
	}
	int kao = 0;
	for (int i = 0; i < res.size() / K; i++) {
		reverse(res.begin() + kao, res.begin() + kao + K);
		kao = kao+K;
	}
	for (int i = 0; i < res.size(); i++) {
		if (i != res.size()-1)
		printf("%05d %d %05d\n",res[i].addr,res[i].data,res[i+1].addr);
		else
		printf("%05d %d -1\n", res[i].addr, res[i].data);

	}
	return 0;
}
 

1075 PAT Judge （25 point(s)）

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 positive integers, N (≤10​4​​), the total number of users, K (≤5), the total number of problems, and M (≤10​5​​), the total number of submissions. It is then assumed that the user id's are 5-digit numbers from 00001 to N, and the problem id's are from 1 to K. The next line contains K positive integers p[i] (i=1, ..., K), where p[i] corresponds to the full mark of the i-th problem. Then M lines follow, each gives the information of a submission in the following format:

user_id problem_id partial_score_obtained

where partial_score_obtained is either −1 if the submission cannot even pass the compiler, or is an integer in the range [0, p[problem_id]]. All the numbers in a line are separated by a space.

Output Specification:

For each test case, you are supposed to output the ranklist in the following format:

rank user_id total_score s[1] ... s[K]

where rank is calculated according to the total_score, and all the users with the same total_score obtain the same rank; and s[i] is the partial score obtained for the i-th problem. If a user has never submitted a solution for a problem, then "-" must be printed at the corresponding position. If a user has submitted several solutions to solve one problem, then the highest score will be counted.

The ranklist must be printed in non-decreasing order of the ranks. For those who have the same rank, users must be sorted in nonincreasing order according to the number of perfectly solved problems. And if there is still a tie, then they must be printed in increasing order of their id's. For those who has never submitted any solution that can pass the compiler, or has never submitted any solution, they must NOT be shown on the ranklist. It is guaranteed that at least one user can be shown on the ranklist.

Sample Input:

7 4 20
20 25 25 30
00002 2 12
00007 4 17
00005 1 19
00007 2 25
00005 1 20
00002 2 2
00005 1 15
00001 1 18
00004 3 25
00002 2 25
00005 3 22
00006 4 -1
00001 2 18
00002 1 20
00004 1 15
00002 4 18
00001 3 4
00001 4 2
00005 2 -1
00004 2 0

Sample Output:

1 00002 63 20 25 - 18
2 00005 42 20 0 22 -
2 00007 42 - 25 - 17
2 00001 42 18 18 4 2
5 00004 40 15 0 25 -

題目大意：模擬一下PAT考試的排名

解題思路：坑死！怎麼搞都過不去最後一個樣例，找了別人的程式碼來交只能，先說下我的想法，其實就是正常人做的模擬，用一個id關聯到結構體陣列，然後最後一個樣例一直超時，然後我發現。。map雖然當key不存在的時候會自動建立一個對應的value，但是這個過程是極為耗時的，所以要在一開始初始化，但是初始化完成後，超時是不超時了，開始WA了。。不知道為什麼，我現在想想可能是因為存在一種很多人，但是沒有一個人能上榜的情況，這樣我的程式碼是跑不了的，當然只是猜測。我也很絕望。

我的程式碼：

#include<iostream>
#include<string.h>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<time.h>
#include<math.h>
#include<set>
#include<list>
#include<climits>
#include<queue>
#include<cstring>
#include<map>
#include<stack>
#include<string>
using namespace std;
int K;
struct stu {
	int id;
	map<int, int> score;
	bool flag=false;
	int solans = 0;
	int total = 0;
	//map<int, bool> ansp;
	int index;
		
	stu() :flag(false){
		for (int i = 0; i < K; i++) {
		score[i + 1] = -1;
	    }
	}
};
int pro[6];
vector<stu> res;
map<int, stu> lt;
bool cmp(stu a, stu b) {
	if (a.total > b.total) {
		return true;
	}
	else if (a.total == b.total) {
		if (a.solans > b.solans) {
			return true;
		}
		else if (a.solans == b.solans) {
			return a.id < b.id;
		}
		else
			return false;
	}
	else {
		return false;
	}
}
int main() {
	int N, M;
	scanf("%d %d %d",&N,&K,&M);
	for (int i = 0; i < N; i++) {
		stu a;
		lt[i + 1] = a;
	}
	for (int i = 0; i < K; i++)scanf("%d", &pro[i]);
	for (int i = 0; i < M; i++) {
		int tmpid;
		int tmpto;
		int tmpsco;
		scanf("%d %d %d", &tmpid, &tmpto, &tmpsco);
			if (tmpsco == -1) {
				lt[tmpid].id = tmpid;
				//lt[tmpid].ansp[tmpto] = true;
				lt[tmpid].score[tmpto] = 0;
			}
			else {
				lt[tmpid].id = tmpid;
				//lt[tmpid].ansp[tmpto] = true;
				if (lt[tmpid].score[tmpto] < tmpsco) {
					//lt[tmpid].total = lt[tmpid].total - lt[tmpid].score[tmpto] + tmpsco;
					lt[tmpid].score[tmpto] = tmpsco;
				}
				if (lt[tmpid].score[tmpto] == pro[tmpto - 1])
					lt[tmpid].solans += 1;
				lt[tmpid].flag = true;
			}
	}
	for (auto i = lt.begin(); i != lt.end(); i++) {
		if (i->second.flag) {
			for (int j = 0; j < K; j++) {
				if(i->second.score[j + 1]!=-1)
				i->second.total += i->second.score[j + 1];
			}
			res.push_back(i->second);
		}
	}
	sort(res.begin(), res.end(), cmp);
	int index;
	res[0].index = 1;
	for (int i = 1; i < res.size(); i++) {
		if (res[i].total < res[i - 1].total) {
			index = i + 1;
			res[i].index = index;
		}
		else {
			res[i].index = index;
		}
	}
	for (int i = 0; i < res.size(); i++) {
		printf("%d %05d %d ", res[i].index,res[i].id,res[i].total);
		for (int j = 1; j <=K; j++) {
			if (res[i].score[j]!=-1) {
				printf("%d", res[i].score[j]);
			}
			else {
				printf("-");
			}
			if (j != K)
				printf(" ");
			else
				printf("\n");
		}
	}
	return 0;
}

別人的滿分程式碼

struct student {
    int id;
    int score[6];
    bool flag = false;
    int solve = 0;
    int total = 0;
}s[10010];

bool cmp(student a, student b) {
    if (a.total != b.total) return a.total > b.total;
    else if (a.solve != b.solve) return a.solve > b.solve;
    else return  a.id < b.id;
}

int main() {
    int n, k, m, i;
    scanf("%d %d %d", &n, &k, &m);
    int p[6];
    for (i = 1; i <= k; i++) {
        scanf("%d", &p[i]);
    }
    for ( i = 1; i <= n; i++) {//我還是不知道，在我輸入id之後，再賦值s[id].id=id為什麼最後一個測試點過不了，必須在這裡初始化
        s[i].id = i;
        memset(s[i].score, -1, sizeof(s[i].score));
     }

    int id, num, temp;
    for (i = 1; i <= m; i++) {
        scanf("%d %d %d", &id, &num, &temp);
        if (temp != -1) s[id].flag = true;
        if (temp == -1 && s[id].score[num] == -1) s[id].score[num] = 0;
        if (temp == p[num] && s[id].score[num] < p[num]) s[id].solve++;//這個語句要在下一個語句之前，不然temp已經存進去了
        if (temp > s[id].score[num]) s[id].score[num] = temp;

    }
    for (i = 1; i <= n; i++) {
        for (int j = 1; j <= k; j++) {
            if (s[i].score[j] != -1 ) s[i].total += s[i].score[j];
        }
    }
    sort(s + 1, s + n + 1, cmp);
    int r = 1;
    for (i = 1; i <= n && s[i].flag == true; i++) {//只輸出能夠輸出的
        if (i > 1 && s[i].total != s[i - 1].total) r = i;
        printf("%d %05d %d", r, s[i].id, s[i].total);
        for (int j = 1; j <= k; j++) {
            if (s[i].score[j] == -1) printf(" -");
            else printf(" %d", s[i].score[j]);
        }
        printf("\n");
    }
    return 0;
}

自閉了。