网易笔试 编程

1 一片1000*1000的草地，初始站在（1,1）(最左上角)，每一秒小易都会横向或纵向到相邻草地吃草(不会越界)，反派超超手上有n个陷阱，第i个陷阱位置（xi,yi），小易一旦进陷阱就会被捕获，为解救小易，请计算小易最少多少秒可能会走入一个陷阱，提前提醒小易

输入描述：第一行：n：超超的陷阱数

   第二行：n个整数xi，表示陷阱横坐标

   第三行：n个整数yi，表示陷阱纵坐标(坐标均在范围内)

#include <iostream>#include <cstdio>#include <cstdlib>#include <string>#include <cstring>#include <queue>#include <limits.h>using namespace std;#define N 1005struct PP {int x;int y;int step;}te, nt;int n;int a[N][N];int dirx[] = { 1,0,-1,0 };//方向int diry[] = { 0,1,0,-1 };int vis[N][N];int inx[N], iny[N];int mi = INT_MAX;int check(int x, int y) //保证不越界、不二次进入{if (x < 1 || x > 1000 || y < 1 || y > 1000 || vis[x][y] == 1) return 0;else return 1;}void bfs() {te.x = 1; te.y = 1; te.step = 0;vis[te.x][te.y] = 1;queue<PP> q;q.push(te);while (q.size() >= 1) {te = q.front();q.pop();if (a[te.x][te.y] == -1) {mi = te.step; return;}nt.step = te.step + 1;for (int i = 0; i < 4; i++) {nt.x = te.x + dirx[i];nt.y = te.y + diry[i];if (check(nt.x, nt.y) == 1){vis[nt.x][nt.y] = 1;q.push(nt);}}}}int main() {//freopen("in.txt","r",stdin);while (scanf("%d", &n) != EOF) {int i;memset(a, 0, sizeof(a));for (i = 0; i < n; i++) {scanf("%d", &inx[i]);}for (i = 0; i < n; i++) {scanf("%d", &iny[i]);a[inx[i]][iny[i]] = -1;}bfs();printf("%d\n", mi);}}

2  “回文串”正读反读均一样的字符串，“level”"noon"；花花有两个字符串A B，现他想将B插入A中使其变成回文串，求有多少种插入方法，B插入位置不一样就是不同方法

如：A="aba" B="b",有4种插入方法“baba” "abba" "abba" "abab"  答案为2

输入描述：

   输入数据共两行；A、B；字符串<100，只包含小写字母

#include <iostream>#include <cstdio>#include <cstdlib>#include <string>#include <cstring>#include <queue>#include <limits.h>using namespace std;#define N 1005string s, t, c;int szs, szt;int ans;int check(string x) {int i = 0, j = x.size() - 1;while (i <= j) {if (x[i] != x[j]) return 0;i++; j--;}return 1;}void fun(int x) {c = s.substr(0, x) + t + s.substr(x);// << x << " " << c << endl;if (check(c) == 1) {ans++;}}int main() {//freopen("in.txt","r",stdin);//while(scanf("%d",&n) != EOF ) {while (cin >> s >> t) {szs = s.size();szt = t.size();ans = 0;int i;for (i = 0; i <= szs; i++) {fun(i);}printf("%d\n", ans);}}

3  二货小易有一个W*H的网格盒子，网格的行标号0~H-1，网格列标号0~W-1，每个格子至多放一块蛋糕，任意两块蛋糕的欧几里距离不能等于2.

(x1,y1),(x2,y2)的欧几里距离：(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)的算术平方根。

小易在网格内能够放多少块蛋糕？

输出描述：输出最多放的蛋糕数；

输入例子：3 2

输出：4

#include <iostream>#include <cstdio>#include <cstdlib>#include <string>#include <cstring>#include <queue>#include <limits.h>using namespace std;#define N 1005int a[N][N];int n,m;int ans = 0;int main() {    //freopen("in.txt","r",stdin);    while(scanf("%d%d",&n,&m) != EOF ) {    //while(cin >> s >> t) {        memset(a,0,sizeof(a));        int i,j,ni,nj;        ans = 0;        for(i = 0;i < n;i++){            for(j = 0;j < m;j++) {                if(a[i][j] == -1) continue;                ans++;                ni = i + 2;                nj = j;                if(ni < n) {                    a[ni][nj] = -1;                }                ni = i;                nj = j + 2;                if(nj < m) {                    a[ni][nj] = -1;                }            }        }        printf("%d\n",ans);    }}