机试重点题-2019
B：数字配对
考察：map容器的应用
#include <iostream> #include <algorithm> #include <map> using namespace std; int n; map<int, int> mp; int main() { cin >> n; for(int i = 0; i<n; i++) { int temp; cin >> temp; mp[temp] ++; } map<int, int> :: iterator itor = mp.begin(); while(itor != mp.end()) { if(itor->second % 2 != 0) { cout << itor->first << endl; break; } itor++; } return 0; } /* 13 2 5 1 3 5 2 5 1 4 1 4 5 1 ouput:3 5 12345 123456789 1234567 123456789 12345 ouput:1234567 */
D：修路的最小代价
考察：最小生成树
Kruskal算法解决：
#include <iostream> #include <algorithm> #include <cstring> using namespace std; const int N = 10010, M = 20010; int n, m; int h[N], e[M], ne[M], idx; int p[N]; struct Edge{ int a, b, w; }edges[M]; void add(int a, int b) { e[idx] = b; ne[idx] = h[a]; h[a] = idx++; } bool cmp(Edge a, Edge b) { return a.w < b.w; } int find(int x) { if(p[x] != x) p[x] = find(p[x]); return p[x]; } int kruskal() { sort(edges, edges + m, cmp); int res = 0, cnt = 0; for(int i = 0; i<m; i++) { int a = edges[i].a; int b = edges[i].b; int w = edges[i].w; a = find(a), b = find(b); if(a != b) { p[a] = b; cnt ++; res += w; } } if(cnt == n - 1) return res; else return -1; } int main() { memset(h, -1, sizeof h); cin >> n >> m; for(int i = 0; i<n; i++) p[i] = i; for(int i = 0; i<m; i++) { int a, b, w; cin >> a >> b >> w; add(a, b), add(b, a); edges[i] = {a, b, w}; } int t = kruskal(); cout << t << endl; return 0; } /* 5 8 0 1 5 0 2 1 0 3 2 1 2 3 2 3 6 4 1 4 4 2 2 4 3 3 ouput: 8 */
E：炮台放置
考察：DFS
#include <iostream> #include <algorithm> #include <cstring> using namespace std; const int N = 1010, M = 1010; char g[N][M]; int n, m; int cnt = 0, max_cnt = 0; void place(int row, int col) { //向上扫描 for(int i = row - 1; i>0; i--) { if(g[i][col] == 'p') return; if(g[i][col] == 'x') break; } //向下扫描 for(int i = row + 1; i<=n; i++) { if(g[i][col] == 'p') return; if(g[i][col] == 'x') break; } //向左扫描 for(int j = col - 1; j>0; j--) { if(g[row][j] == 'p') return; if(g[row][j] == 'x') break; } //向右扫描 for(int j = col + 1; j<=m; j++){ if(g[row][j] == 'p') return; if(g[row][j] == 'x') break; } g[row][col] = 'p'; cnt ++; //可放炮台数加1 max_cnt = max(max_cnt, cnt);//更新最大炮台数 for(int i = 1; i<=n; i++) for(int j = 1; j<=m; j++) if(g[i][j] == '.') place(i, j); //只要是个.就尝试放炮 g[row][col] = '.'; //恢复现场，进行下一轮放置 cnt --; } int main() { cin >> n >> m; for(int i = 1; i<=n; i++) for(int j = 1; j<=m; j++) cin >> g[i][j]; for(int i = 1; i<=n; i++) for(int j = 1; j<=m; j++) if(g[i][j] == '.') place(i, j); cout << max_cnt << endl; return 0; } /* 4 4 .x.. .... x... .... 2 2 .x x. 5 5 ..... .x..x ..x.. ..... .x... */
F：饭卡
HDU2546 原题
考察：贪心 + 背包
#include <iostream> #include <algorithm> #include <cstring> using namespace std; const int N = 1010, M = 1010; int n, m, k;//菜数量，余额，输入组数 int f[N][M]; //f[i][j]：在余额为j的情况下，从前i道菜里购买所能花费的最大方案 int price[N]; int main() { while(cin >> n && n != 0) { for(int i = 1; i<=n; i++) cin >> price[i]; cin >> m; if(m < 5) { cout << m << endl; //钱不够，余额还是m continue; } sort(price, price + n + 1); //注意题设条件，每购买一件商品之前先判断有没有五块钱 int max_price = price[n]; m -= 5; //先花5元钱买最贵的菜，根据题目条件一定可以买成功 //剩余的钱再用01背包解决 for(int i = 1; i<=n-1; i++){ for(int j = 5; j<=m; j++) { f[i][j] = f[i - 1][j]; if(j >= price[i]) f[i][j] = max(f[i][j], f[i - 1][j - price[i]] + price[i]); } } //之前m花了5元钱买了最贵的菜要加回来 //再减去买的最贵的菜价，再减去从前n-1道菜里用"m"(m-5)买菜的价格 int remain = m + 5 - price[n] - f[n-1][m]; cout << remain << endl; } return 0; } /* 1 50 5 10 1 2 3 2 1 1 2 3 2 1 50 0 ouput: -45 32 */
