摘要： 这道题放在 DFS 里不太合适，我刚开始就想到了 DP 却没写，DFS 超时后看了题解才发现就是记忆化搜索（白书认为记忆化就是DP）；1TLE 2WA，把 lens 写成了 lent 后来才发现的。View Code # include <cstdio># include <cstring># define N 200 + 10bool finished, f[N][N];int lens, lent;char s[N], t[N], g[2 * N];void dfs(int p, int q, int cnt){ if (finished) return; if ( 阅读全文

摘要： 不含回溯，看了标程后 AC 的，标程的代码很短，处理相同解的方法比较好。View Code # include <cstdio># define N 12 + 5bool find;int t, n, a[N], solu[N];void dfs(int sum, int p, int cnt){ if (sum > t) return ; if (sum == t) { find = true; printf("%d", solu[0]); for (int i = 1; i < cnt; ++i) print... 阅读全文

摘要： 有向图的DFS，不能包含回溯，会爆栈的。View Code # include <cstdio># include <cstring># define N 26bool finished, vis[N];char g[N][N];void dfs(int u){ for (int i = 0; i < N; ++i) { if (u != i && g[u][i]) { if (vis[i]) return; if (i == 'm'-'a') {finished = true; return ;} ... 阅读全文

摘要： View Code # include <cstdio># include <cstring># define N 10 + 5int r, c, s, d[N][N];char grid[N][N];void init(void){ scanf("%d", &s); for (int i = 1; i <= r; ++i) { scanf("%s", grid[i]+1); memset(d[i]+1, 0, sizeof(d[i][0])*c); }}void solve(void){ int x, y, t; 阅读全文

摘要： View Code # include <cstdio># include <cstring># define N 20 + 5char ptab[25] = {0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0};int n, solu[N];bool vis[N];void dfs(int cnt){ if (cnt... 阅读全文

摘要： DFS 题目，剪枝比较重要，这里使用的是把重复的方块只记录一次，用 num[] 保存它的数目。# include <cstdio># include <cstring># define N 25 + 2bool finished;int n, m, t[N][4], num[N], ans[N];void dfs(int cnt){ if (cnt == n*n) {finished = true; return ;} int x = cnt/n + 1, y = cnt%n + 1; int left = cnt, top = cnt+1-n; for (int i 阅读全文