摘要： 这道题暴力秒过，以前在ZOJ上见过，当时不会，还以为不剪枝铁定超呢，结果纯暴力就过了。# include <cstdio># include <cstring># define N 4 + 3int n, ans;char g[N][N];bool check(int x, int y){ int i = x; if (g[x][y] != '.') return false; while (i <= n) { if (g[i][y] == 'X') break; else if (g[i][y] == (i-1)*n+y) ret 阅读全文

摘要： 这道题的剪枝：奇偶性，曼哈顿距离类型的。不剪会超时。# include <cstdio># include <cstring>using namespace std;# define N 7 + 3# define ABS(x) (((x)>0)?(x):(-(x)))int si, sj, gi, gj;int n, m, T;char g[N][N];bool vis[N][N];bool finished;const int dir[][2] = {{0,1}, {1,0}, {0,-1}, {-1,0}};void dfs(int x, int y, in 阅读全文

摘要： 这道题如果用 DFS 必须剪枝：剪掉重复的组合。DFS 枚举了所有的排列，实际上这道题只需要组合。View Code # include <cstdio># include <cstring># define N 20 + 5int ans;int n, k, vMax, v[N], w[N];bool vis[N];void dfs(int cur, int sum, int cnt, int weight){ if (weight > vMax) return ; if (cnt == k) ans = (sum > ans ? sum : ans); 阅读全文

摘要： HINT 误导人，水题。View Code # include <cstdio># include <cstring># define N 15 + 5int n, ans;int f[N][N];bool vis[N];void dfs(int cur, int cnt, int t){ if (cnt > ans) ans = cnt; for (int i = 0; i < n; ++i) if (vis[i] == false && f[cur][i] >= t) { vis[i] = true; dfs(i, cnt+1, f 阅读全文

摘要： DFS 其实会超时的，打表。# include <cstdio># include <cstring># define N 10 + 5int n, ans;int solu[N];bool vis[N];void dfs(int cnt){ if (cnt == n) { ++ans; return ; } bool ok; for (int i = 1; i <= n; ++i) if (vis[i] == false) { ok = true; for (int j = 1; j <= cnt; ... 阅读全文

摘要： 注意：标号后面是两个空格。# include <cstdio># include <cstring># include <algorithm>using namespace std;int n, m, adj[21][3];bool vis[21];int solu[22];void dfs(int cur, int cnt){ int t; for (int i = 0; i < 3; ++i) { t = adj[cur][i]; if (cnt == 20 && t == m) { printf("%d: ... 阅读全文

摘要： 这道题目有问题，4WA ：注意：对于每组测试，Lele都是在站点0拉上乘客的。最后看了题解，改了初始状态 AC 的。View Code # include <cstdio># include <cstdlib># include <cstring># define N 30 + 5int n, k;int g[N][N], des[N], min;bool vis[N];int cmp(const void *x, const void *y){ return *(int*)x - *(int*)y;}void dfs(int cnt, int u, int 阅读全文