codeforces 589H - Tourist Guide（dfs）

基础的搜索BFS和DFS，自己找题切吧… 高级搜索的题集就在下面，自己看着办吧… 努力爆搜，努力剪枝吧~~~ 【Level 1】

＿＿＿＿＿＿ 好蠢，竟然没看出来这道题的不同之处，以为就是个搜 然后样例什么的都过了… 结果显然 WA… 然后才发现，这道题应该是 tsp 问题。 解法是先跑一遍 bfs， 对于所有的脏点和起点，得到每两个点之间的距离。

dfs 1A 代码实现 1/************************************************************************* 2 > File Name: code/whust/#9/K.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年08月05日 星期三 15时02分30秒 6 ************************************************************************/ 7 8#include<iostream> 9#include<iomanip> 10#include<cstdio> 11#include<algorithm> 12#include<cmath> 13#include<cstring> 14#include<string> 15#include<map> 16#include<set> 17#include<queue> 18#include<vector> 19#include<stack> 20#define y0 abc111qqz 21#define y1 hust111qqz 22#define yn hez111qqz 23#define j1 cute111qqz 24#define tm crazy111qqz 25#define lr dying111qqz 26using namespace std; 27#define REP(i, n) for (int i=0;i<int(n);++i) 28typedef long long LL; 29typedef unsigned long long ULL; 30const int inf = 0x7fffffff; 31int n ,m,ans; 32int d[50]; 33int on[50],off[50]; 34int u[50],v[50]; 35 36bool ok(int x,int y) 37{ 38 if ( !d[x] && !d[y] && on[x] == off[x] && on[y] == off[y] ) 39 return true; 40 if ( !d[x] && d[y] && on[x] == off[x] ) 41 return true; 42 if ( d[x] && !d[y] && on[y] == off[y] ) 43 return true; 44 if ( d[x] && d[y] ) 45 return true; 46 return false; 47} 48 49void dfs ( int i) 50{ 51 if (i==m) 52 { 53 ans++; 54 return; 55 } 56 int x = u[i]; 57 int y = v[i]; 58 d[x]--; 59 d[y]--; 60 on[x]++; 61 on[y]++; 62 if (ok(x,y)) 63 dfs (i+1); 64 on[x]--;on[y]--; 65 off[y]++;off[x]++; 66 if ( ok( x,y)) 67 dfs (i+1); 68 off[y]--; 69 off[x]--; 70 d[x]++; 71 d[y]++; 72} 73 74int main ( ) 75{ 76 int T; 77 cin>>T; 78 while ( T-- ) 79 { 80 ans = 0; 81 scanf ("%d %d",&n,&m); 82 memset (d,0,sizeof(d)); 83 memset (on,0,sizeof(on)); 84 memset (off,0,sizeof(off)); 85 for ( int i = 0 ; i < m ; i++ ) 86 { 87 scanf ("%d%d",&u[i],&v[i]); 88 d[u[i]]++; 89 d[v[i]]++; 90 } 91 dfs (0); 92 printf ( "%d\n" , ans ); 93 } 94}

“… so forward this to ten other people, to prove that you believe the emperor has 题意是说发短信，每个人只会给一个人发，问从哪个人开始发，能传到的人最多 思路是每个人开始做一遍dfs… 毫无意外的TLE了 一个容易想到的剪枝是，如果在第i次之前的路径上的点，在之后以它作为起点遍历一定不优.

Oil Deposits # **Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17683 Accepted Submission(s): 10172 ** Problem Description The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.