题意：n个点，给定起点和终点，可以每次可以走一格或两格，走一格则需要一个代价，每个格子只能走一次，问从起点到终点并经过每一个点的最小代价
思路：这题我没看出什么道理，先打了个暴力，结果发现了个相当坑的规律，，然后就过了。
1 #include <iostream> 2 #include <cstdio> 3 #include <fstream> 4 #include <algorithm> 5 #include <cmath> 6 #include <deque> 7 #include <vector> 8 #include <queue> 9 #include <string> 10 #include <cstring> 11 #include <map> 12 #include <stack> 13 #include <set> 14 #define LL long long 15 #define eps 1e-8 16 #define INF 0x3f3f3f3f 17 #define MAXN 200005 18 using namespace std; 19 int main() 20 { 21 #ifndef ONLINE_JUDGE 22 freopen("in.txt", "r", stdin); 23 freopen("out.txt", "w", stdout); 24 #endif // OPEN_FILE 25 int n, s, t, o; 26 while(~scanf("%d%d%d", &n, &s, &t)){ 27 //printf("%d %d %d %d ", n, s, t, o); 28 int ans; 29 if(s == t){ 30 ans = 2; 31 } 32 else{ 33 if(s > t){ 34 swap(s, t); 35 } 36 if(t - s == 1){ 37 if(s != 1 && t != n){ 38 ans = -1; 39 } 40 else{ 41 ans = 1; 42 } 43 } 44 else{ 45 int tt = n -s + 1; 46 int ss = n - t + 1; 47 if(ss < s){ 48 s = ss; 49 t = tt; 50 } 51 int p; 52 if(s == 1){ 53 p = t - s - 1; 54 ans = 1 + p / 3 + (p % 3); 55 } 56 else{ 57 p = t - s - 2; 58 ans = 2 + p / 3 + (p % 3); 59 } 60 } 61 } 62 if(s == 1 && t == n && n % 3 == 1){ 63 ans -= 2; 64 } 65 printf("%d ", ans); 66 } 67 /*while(~scanf("%d%d%d", &n, &s, &t)){ 68 memset(vis, 0, sizeof(vis)); 69 vis[s] = true; 70 vis[t] = true; 71 ans = INF; 72 dfs(s, 0, 0); 73 if(ans == INF){ 74 printf("%d %d %d -1 ", n, s, t); 75 continue; 76 } 77 printf("%d %d %d %d ", n, s, t, ans); 78 }*/ 79 }
下面是暴力的
1 #include <iostream> 2 #include <cstdio> 3 #include <fstream> 4 #include <algorithm> 5 #include <cmath> 6 #include <deque> 7 #include <vector> 8 #include <queue> 9 #include <string> 10 #include <cstring> 11 #include <map> 12 #include <stack> 13 #include <set> 14 #define LL long long 15 #define eps 1e-8 16 #define INF 0x3f3f3f3f 17 #define MAXN 200005 18 int n, s, t; 19 int ans; 20 bool vis[MAXN]; 21 using namespace std; 22 void dfs(int x, int cnt, int cost){ 23 if(cnt == n - 2){ 24 if(abs(x - t) == 1){ 25 ans = min(ans, cost + 1); 26 } 27 else if(abs(x - t) == 2){ 28 ans = min(ans, cost); 29 } 30 else{ 31 return; 32 } 33 } 34 int y; 35 for(int i = 1; i <= 2; i++){ 36 int p = i == 1 ? 1 : 0; 37 y = x + i; 38 if(y >= 1 && y <= n && !vis[y]){ 39 vis[y] = true; 40 dfs(y, cnt + 1, cost + p); 41 vis[y] = false; 42 } 43 y = x - i; 44 if(y >= 1 && y <= n && !vis[y]){ 45 vis[y] = true; 46 dfs(y, cnt + 1, cost + p); 47 vis[y] = false; 48 } 49 } 50 } 51 int main() 52 { 53 #ifndef ONLINE_JUDGE 54 freopen("in.txt", "r", stdin); 55 //freopen("out.txt", "w", stdout); 56 #endif // OPEN_FILE 57 58 while(~scanf("%d%d%d", &n, &s, &t)){ 59 memset(vis, 0, sizeof(vis)); 60 vis[s] = true; 61 vis[t] = true; 62 ans = INF; 63 dfs(s, 0, 0); 64 if(ans == INF){ 65 printf("-1 "); 66 return 0; 67 } 68 printf("%d ", ans); 69 } 70 }