随笔分类 -  回溯

摘要：DFS，要求输出字典序最小的，注意扩展方向。# include <cstdio># include <cstring># define N 26 + 5const int dx[] = {-1, 1,-2, 2,-2, 2,-1, 1};const int dy[] = {-2,-2,-1,-1, 1, 1, 2, 2};int p, q, cnt;bool finished;char solu[N][2];char vis[N][N];void dfs(int x, int y){ solu[cnt][0] = x, solu[cnt][1] = y; if (cnt 阅读全文

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

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 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

摘要：dfs；按照 Staginner 大牛的方法写的，大致思路是：刚开始所有点没有着色，且最终结果至少有一个点被着黑色（一个点时直接着黑色，多个点时，可以任选一个点为黑色，其余点全为白色）；枚举这个黑色的点，并且把与它相邻的点都着白色，剩下的可以看作是一个相同的子问题了，因为剩下的点都不与这个黑色的点相邻。最优解满足：每个白色的点至少与一个黑色的点相邻（如果这个点相邻的都是白色，可以把它改为黑色），且每个黑色的点周围都是白色。# include <stdio.h># include <string.h># define N 105int n, m, ans;char g[N 阅读全文

摘要：回溯，看了lrj的白书后写的；一提交，TLE。# include <stdio.h># define MAXN 15int n, ans;char vis[3][MAXN*2+1];void search(int cur);int main(){ while (~scanf("%d", &n)) { if (!n) break; ans = 0; memset(vis, 0, sizeof(vis)); search(0); printf("%d\n", ans)... 阅读全文