牛客练习赛24题解（搜索，DP）
A题，C题不讲，基础题（但是我要抨击一下这次比赛，卡cin,cout，卡的太狠了，根本就不让过的那种，QAQ）
链接：https://www.nowcoder.com/acm/contest/157/B
来源：牛客网
凤 凰
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld
题目描述
凤凰于飞，翙翙其羽，亦集爰止。
——《诗经·卷阿》
传说，凤凰是百鸟之王。有一天，凤凰要召开百鸟大会，百鸟国是一个由n个节点组成的树，每个节点有一只鸟，开会的节点定在1号节点。每只鸟可以花费1s通过一条边，由于每根树枝(边)的载重有限，只允许一只鸟同时通过。作为会议的策划师，HtBest想知道百鸟国的所有鸟在1点集合最少需要多少秒。
输入描述:
第一行有一个正整数n，表示百鸟国节点个数。
接下来n-1行，第i行两个正整数a
i
,b
i
用空格隔开，表示树上节点a
i
,b
i
之间有一条边。
输出描述:
第一行一个整数，表示集合最少需要的时间。
输入例子:
3
1 2
2 3
输出例子:
2
-->
备注:
对于100%的测试数据：
1 ≤ n ≤ 1000000
数据量较大，注意使用更快的输入输出方式。
解析：有人是用搜索过的，我的第一想法就是并查集了（收获了并查集大佬的写法，也是蛮开心的，然后要注意的就是不要连接与起点有关的任何点）
，这样对于任何分枝就都能连接起来，起到+1的作用，
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. **/
#include <iostream>
#include <algorithm>
#include <vector>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
#include <stdio.h>
using namespace std;
#define FFI(a, b) for(int i = a; i < b; i++)
#define FFJ(a, b) for(int j = a; j < b; j++)
#define FFK(a, b) for(int k = a; k < b; k++)
#define RR(a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define SCC(a, b) scanf("%d%d", &a, &b)
#define SCCC(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define PR(x) printf("%d\n", x)
#define PRR(a, b) printf("%d%d\n", a, b);
#define INF 0x3f3f3f3f
#define MAX 1314520
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP;
/*******************************************************************************************************************************************************/
//QAQ，我就说是并查集，为什么不相信我，QAQ
int pre[MAX], num[MAX];
//int Find(int t){
// int ret = t;
// while(pre[t] != t) t = pre[t];
//
// while(t != pre[ret]){
// int temp = pre[ret];
// pre[ret] = t;
// ret = temp;
// }
// return t;
//}
int Find(int t){return t == pre[t] ? t : pre[t] = Find(pre[t]);} //比我写的牛逼多了，orz
int main(void){
#ifdef Empirefree
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie(); cout.tie();
int n, a, b, x, y, ans;
cin>>n;ME(num, );ans = ;
FFI(, n + ) pre[i] = i;
FFI(, n){
cin>>x>>y;
a = Find(x);b = Find(y);
if(x != && y != ) pre[a] = b; //这里最重要，除去原点，防止连接成直线
}
FFI(, n + ) num[Find(i)]++;
cout<< *max_element(num + , num + n + )<<endl;
return EXIT_SUCCESS;
}
Empirefree.top
链接：https://www.nowcoder.com/acm/contest/157/D
来源：牛客网
插排树
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
题目描述
一年一度的山东省oi夏令营又开始了，每到这个季节，山东的oier们都会欢聚这里，一起学（tuí）习（feì）。当然，为了能更加愉快地学（tuí）习（feì），就少不了要自带电脑，用电便开始成了一种问题，于是便有一种神奇的数据结构诞生了！这就是山东省oi专用数据结构——插排树（如图）
小K为了能更好的学（tuí）习（feì），所以他想尽量的往后做，所以现在请你帮帮他，他最远可以离讲台多远。
已知插排树的根节点在讲台上，有且仅有一个根节点（根节点入度为0），最远距离即所有插排的长度，小K电脑线的长度忽略不计
本题良心大样例下载地址： https://kench.co/tree.zip
输入描述:
第一行一个整数n表示有n个节点
然后n-1行，每行三个整数a,b,c，表示插排a是接在插排b上的，插排a的长度为c
输出描述:
一个整数n表示最远距离
输入例子:
9
2 1 2
3 1 2
4 1 1
5 2 3
6 2 1
7 3 1
8 3 4
9 7 5
输出例子:
8
-->
备注:
对于30%的数据 n<233
对于70%的数据 n<2333 对于100%的数据 n<50000 c小于20 a,b小于等于n
解析：对于每个点，暴力搜索一下，每次保留最大值
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. **/
#include <iostream>
#include <algorithm>
#include <vector>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
#include <stdio.h>
using namespace std;
#define FFI(a, b) for(int i = a; i < b; i++)
#define FFJ(a, b) for(int j = a; j < b; j++)
#define FFK(a, b) for(int k = a; k < b; k++)
#define RR(a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define SCC(a, b) scanf("%d%d", &a, &b)
#define SCCC(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define PR(x) printf("%d\n", x)
#define PRR(a, b) printf("%d%d\n", a, b);
#define INF 0x3f3f3f3f
#define MAX 55555
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
/*******************************************************************************************************************************************************/
vector<Author> VP[MAX];
int dis[MAX], ans;
void DFS(int index, int sum){
int len = VP[index].size();
ans = max(ans, sum);
FFI(, len)DFS(VP[index][i].first, sum + VP[index][i].second);
}
int main(void){
#ifdef Empirefree
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie(); cout.tie();
int n, a, b, c;
SC(n);ME(dis, );ans = ;
FFI(, n){
SCCC(a, b, c);
//b插在a上，且从b开始搜索
VP[b].push_back(make_pair(a, c));
dis[a] = ;
}
FFI(, n + )if(!dis[i])DFS(i, );
PR(ans);
return EXIT_SUCCESS;
}
Empirefree.top
链接：https://www.nowcoder.com/acm/contest/157/E
来源：牛客网
青蛙
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
题目描述
有一只可爱的老青蛙，在路的另一端发现了一个黑的东西，想过去一探究竟。于是便开始踏上了旅途
一直这个小路上有很多的隧道，从隧道的a进入，会从b出来，但是隧道不可以反向走。
这只青蛙因为太老了，所以很懒，现在想请你帮帮慢，问他最少需要几步才可以到达对面。
将小径看作一条数轴，青蛙初始在0上，这只青蛙可以向前跳也可以向后跳，但每次只能跳一格，每跳一格记作一步，从隧道进到隧道出算做一步。
输入描述:
第一行两个数m,n；表示黑色物品在数轴m点上，数轴上总共有n个隧道
接下来n行，每行a，b两个数，表示从a进会从b出 10 <= m,n <= 233 0<a,b<=m
输出描述:
一个数ans表示最小步数
输入例子:
16 4
2 10
8 15
12 5
13 6
输出例子:
7
-->
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. **/
#include <iostream>
#include <algorithm>
#include <vector>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
#include <stdio.h>
using namespace std;
#define FFI(a, b) for(int i = a; i < b; i++)
#define FFJ(a, b) for(int j = a; j < b; j++)
#define FFK(a, b) for(int k = a; k < b; k++)
#define RR(a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define SCC(a, b) scanf("%d%d", &a, &b)
#define SCCC(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define PR(x) printf("%d\n", x)
#define PRR(a, b) printf("%d%d\n", a, b);
#define INF 0x3f3f3f3f
#define MAX 234
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP;
/*******************************************************************************************************************************************************/
//类似一维上的传送门题
int vis[MAX];
int dis[MAX];
int n;
int dir[] = {-, };
struct Node{
int x, time;
bool operator < (const Node& a)const{return time > a.time;}
}bn, en, temp1, temp2, cnt;
int BFS(){
priority_queue<Node> q;ME(dis, );
dis[] = ;q.push(bn);
while(!q.empty()){
temp1= q.top();q.pop();
if(temp1.x == en.x) return temp1.time;
FFI(, ){
temp2.x = temp1.x + dir[i];
temp2.time = temp1.time + ;
if(temp2.x >= && temp2.x <= n && !dis[temp2.x]){
dis[temp2.x] = ;
q.push(temp2);
}
}
if(vis[temp1.x]){
cnt.x = vis[temp1.x];
cnt.time = temp1.time + ;
if(cnt.x >= && cnt.x <= n && !dis[cnt.x]) q.push(cnt);
}
}
return ;
}
int main(void){
#ifdef Empirefree
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie(); cout.tie();
int m, a, b;
cin>>n>>m;ME(vis, );
FFI(, m){
cin>>a>>b;
vis[a] = b;
}
bn.x = ;bn.time = ;en.x = n;
cout<<BFS()<<endl;
return EXIT_SUCCESS;
}
Empirefree.top的BFS
/**
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. **/
#include <iostream>
#include <algorithm>
#include <vector>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
#include <stdio.h>
using namespace std;
#define FFI(a, b) for(int i = a; i < b; i++)
#define FFJ(a, b) for(int j = a; j < b; j++)
#define FFK(a, b) for(int k = a; k < b; k++)
#define RR(a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define SCC(a, b) scanf("%d%d", &a, &b)
#define SCCC(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define PR(x) printf("%d\n", x)
#define PRR(a, b) printf("%d%d\n", a, b);
#define INF 0x3f3f3f3f
#define MAX 234
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP;
/*******************************************************************************************************************************************************/
vector<int> vec[MAX];
int dis[MAX], n;
void DFS(int index, int sum){
int len;
dis[index] = sum;
len = vec[index].size();
if(index <= n && dis[index + ] > sum + )DFS(index + , sum + );
if(index - >= && dis[index - ] > sum + )DFS(index - , sum + );
FFI(, len)if(dis[vec[index][i]] > sum + )DFS(vec[index][i], sum + );
}
int main(void){
#ifdef Empirefree
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie(); cout.tie();
int m, a, b;
cin>>n>>m;ME(dis, INF);
FFI(, m){
cin>>a>>b;
vec[a].push_back(b);
}
DFS(, );
cout<<dis[n]<<endl;
return EXIT_SUCCESS;
}
Empirefree.top的DFS
/**
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.:J: :; .,:K6BQS7:.,.,.
:r7Ji;r7;::;;.
. **/
#include <iostream>
#include <algorithm>
#include <vector>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
#include <stdio.h>
using namespace std;
#define FFI(a, b) for(int i = a; i < b; i++)
#define FFJ(a, b) for(int j = a; j < b; j++)
#define FFK(a, b) for(int k = a; k < b; k++)
#define RR(a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define SCC(a, b) scanf("%d%d", &a, &b)
#define SCCC(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define SCCCC(a, b, c, d) scanf("%d%d%d%d", &a, &b, &c, &d)
#define PR(x) printf("%d\n", x)
#define PRR(a, b) printf("%d%d\n", a, b);
#define INF 0x3f3f3f3f
#define MAX 333
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP;
/*******************************************************************************************************************************************************/
int dis[MAX][MAX];char matrix[MAX][MAX];
struct Node{
int x;
int y;
int time;
}bn, en, temp1, temp2;
vector<Node> vec[MAX][MAX];
int n, m, q;
bool Check(int x, int y){
if(x >= && x < n && y >= && y < m && matrix[x][y] != '#') return true;
else return false;
}
void DFS(int x,int y, int sum){
int len;
dis[x][y] = sum;
if(Check(x + , y) && dis[x + ][y] > sum + ) DFS(x + , y, sum + );
if(Check(x - , y) && dis[x - ][y] > sum + ) DFS(x - , y, sum + );
if(Check(x, y + ) && dis[x][y + ] > sum + ) DFS(x, y + , sum + );
if(Check(x, y - ) && dis[x][y - ] > sum + ) DFS(x, y - , sum + );
len = vec[x][y].size();
FFI(, len){
if(Check(vec[x][y][i].x, vec[x][y][i].y) && dis[vec[x][y][i].x][vec[x][y][i].y] > sum + )
DFS(vec[x][y][i].x, vec[x][y][i].y, sum + );
}
}
void Clear(){
FFI(, n)
FFJ(, m)
vec[i][j].clear();
}
int main(void){
#ifdef Empirefree
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie(); cout.tie();
int a, b, c, d;
while(SCCC(n, m, q)){
ME(dis, INF);Clear();
FFI(, n){
scanf("%s", matrix[i]);
FFJ(, m){
if(matrix[i][j] == 'S') bn.x = i,bn.y = j;
if(matrix[i][j] == 'T') en.x = i,en.y = j;
}
}
FFI(, q){
SCCCC(a, b, c, d);
temp1.x = c;temp1.y = d;
vec[a][b].push_back(temp1);
}
DFS(bn.x, bn.y, );
if(dis[en.x][en.y] == INF) PR(-);
else PR(dis[en.x][en.y]);
}
return EXIT_SUCCESS;
}
maze的DFS超时代码
链接：https://www.nowcoder.com/acm/contest/157/F
来源：牛客网
三轮
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld
题目描述
解析：完全背包，QAQ
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. **/
#include <iostream>
#include <algorithm>
#include <vector>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
#include <stdio.h>
using namespace std;
#define FFI(a, b) for(int i = a; i < b; i++)
#define FFJ(a, b) for(int j = a; j < b; j++)
#define FFK(a, b) for(int k = a; k < b; k++)
#define RR(a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define SCC(a, b) scanf("%d%d", &a, &b)
#define SCCC(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define PR(x) printf("%d\n", x)
#define PRR(a, b) printf("%d%d\n", a, b);
#define INF 0x3f3f3f3f
#define MAX 52000
#define MOD 19260817
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP;
int read()
{
int x=,f=; char ch;
while (ch=getchar(),ch<''||ch>'') if (ch=='-') f=-;
while (x=x*+ch-'',ch=getchar(),ch>=''&&ch<='');
return x*f;
}
/*******************************************************************************************************************************************************/
LL bag[MAX];
int main(void){
#ifdef Empirefree
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie(); cout.tie();
LL n, m, a, ans;
bag[] = ;ans = ;
n = read(); m = read();
FFI(, n){
a = read();
FFJ(a, m + )
bag[j] = (bag[j] + bag[j - a]) % MOD;
}
FFI(, m + ) ans = (ans + bag[i]) % MOD;
printf("%lld\n", ans);
return EXIT_SUCCESS;
}
Empirefree.top
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