1052 Linked List Sorting （25 point(s)）
A linked list consists of a series of structures, which are not necessarily adjacent in memory. We assume that each structure contains an integer key and a Next pointer to the next structure. Now given a linked list, you are supposed to sort the structures according to their key values in increasing order.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive N (<105) and an address of the head node, where N is the total number of nodes in memory and the address of a node is a 5-digit positive integer. NULL is represented by −1.
Then N lines follow, each describes a node in the format:
Address Key Next
where Address is the address of the node in memory, Key is an integer in [−105,105], and Next is the address of the next node. It is guaranteed that all the keys are distinct and there is no cycle in the linked list starting from the head node.
Output Specification:
For each test case, the output format is the same as that of the input, where N is the total number of nodes in the list and all the nodes must be sorted order.
Sample Input:
5 00001
11111 100 -1
00001 0 22222
33333 100000 11111
12345 -1 33333
22222 1000 12345
Sample Output:
5 12345
12345 -1 00001
00001 0 11111
11111 100 22222
22222 1000 33333
33333 100000 -1题目大意：给一堆节点的结构，让你输出一个链表，且链表的key值从小到大排列
解题思路：一开始想当然以为节点都有效，后来发现节点有些是不存在的，然后写了一下超时，大概是cin和cout的问题，然后我就直接抄了别人了。就要注意没有节点的情况，当然这种题目不可能真的用链表的，一般都是哈希搞一下。
代码如下：
#include<iostream>
#include<string>
#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>
using namespace std;
struct NODE {
int address, key, next;
bool flag;
}node[100000];
int cmp1(NODE a, NODE b) {
return !a.flag || !b.flag ? a.flag > b.flag : a.key < b.key;
}
int main() {
int n, cnt = 0, s, a, b, c;
scanf("%d%d", &n, &s);
for(int i = 0; i < n; i++) {
scanf("%d%d%d", &a, &b, &c);
node[a] = {a, b, c, false};
}
for(int i = s; i != -1; i = node[i].next) {
node[i].flag = true;
cnt++;
}
if(cnt == 0) {
printf("0 -1");
} else {
sort(node, node + 100000, cmp1);
printf("%d %05d\n", cnt, node[0].address);
for(int i = 0; i < cnt; i++) {
printf("%05d %d ", node[i].address, node[i].key);
if(i != cnt - 1)
printf("%05d\n", node[i + 1].address);
else
printf("-1\n");
}
}
return 0;
}1054 The Dominant Color （20 point(s)）
Behind the scenes in the computer's memory, color is always talked about as a series of 24 bits of information for each pixel. In an image, the color with the largest proportional area is called the dominant color. A strictly dominant color takes more than half of the total area. Now given an image of resolution M by N (for example, 800×600), you are supposed to point out the strictly dominant color.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive numbers: M (≤800) and N (≤600) which are the resolutions of the image. Then N lines follow, each contains M digital colors in the range [0,224). It is guaranteed that the strictly dominant color exists for each input image. All the numbers in a line are separated by a space.
Output Specification:
For each test case, simply print the dominant color in a line.
Sample Input:
5 3
0 0 255 16777215 24
24 24 0 0 24
24 0 24 24 24
Sample Output:
24题目大意：题目花里胡哨，实际上就是给个矩阵，求矩阵内出现次数最多的数。
解题思路：用map存一下每个数出现的次数，然后遍历一下就可以了。
#include<iostream>
#include<string>
#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>
using namespace std;
map<int, int> a;
int main()
{
int n, m;
scanf("%d %d", &m, &n);
for (int i = 0; i < n; i++) {
for (int i = 0; i < m; i++) {
int q;
scanf("%d", &q);
a[q] += 1;
}
}
int max = 0;
int res;
for (auto i = a.begin(); i != a.end(); i++) {
if (i->second > max) {
max = i->second;
res = i->first;
}
}
printf("%d\n", res);
return 0;
}