[歐拉回路] poj 1300 Door Man
阿新 • • 發佈:2017-07-11
linker center || 是否 connect sep cto -m vector
題目鏈接:
http://poj.org/problem?id=1300
Door Man
Description You are a butler in a large mansion. This mansion has so many rooms that they are merely referred to by number (room 0, 1, 2, 3, etc...). Your master is a particularly absent-minded lout and continually leaves doors open throughout a particular floor of the house. Over the years, you have mastered the art of traveling in a single path through the sloppy rooms and closing the doors behind you. Your biggest problem is determining whether it is possible to find a path through the sloppy rooms where you:
In this problem, you are given a list of rooms and open doors between them (along with a starting room). It is not needed to determine a route, only if one is possible. Input Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets.A single data set has 3 components:
Following the final data set will be a single line, "ENDOFINPUT". Note that there will be no more than 100 doors in any single data set. Output For each data set, there will be exactly one line of output. If it is possible for the butler (by following the rules in the introduction) to walk into his chambers and close the final open door behind him, print a line "YES X", where X is the number of doors he closed. Otherwise, print "NO".Sample Input START 1 2 1 END START 0 5 1 2 2 3 3 4 4 END START 0 10 1 9 2 3 4 5 6 7 8 9 END ENDOFINPUT Sample Output YES 1 NO YES 10 Source field=source&key=South+Central+USA+2002" style="text-decoration:none">South Central USA 2002 |
[Submit] [Go Back] [Status]
[ problem_id=1300" style="text-decoration:none">Discuss]
有n個房間,房間之間通過門連接,知道門連接的房間情況。求從m號房間是否能經過全部的門一次。而且回到0號門。
解題思路:
把房間看成節點,門看成邊,由題意知是一個連通圖,然後推斷是否存在從m到0的歐拉通路。
統計各點的度數就可以。
代碼:
//#include<CSpreadSheet.h> #include<iostream> #include<cmath> #include<cstdio> #include<sstream> #include<cstdlib> #include<string> #include<string.h> #include<cstring> #include<algorithm> #include<vector> #include<map> #include<set> #include<stack> #include<list> #include<queue> #include<ctime> #include<bitset> #include<cmath> #define eps 1e-6 #define INF 0x3f3f3f3f #define PI acos(-1.0) #define ll __int64 #define LL long long #define lson l,m,(rt<<1) #define rson m+1,r,(rt<<1)|1 #define M 1000000007 //#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; #define Maxn 22 int nu[Maxn],n,m; char temp[Maxn]; int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); while(scanf("%s",temp)) { if(temp[3]=='O') break; scanf("%d%d",&m,&n); memset(nu,0,sizeof(nu)); getchar(); int ans=0; for(int i=0;i<n;i++) { int la=0; char c; while((c=getchar())!='\n') { if(c==' ') { ans++; nu[i]++; nu[la]++; la=0; while((c=getchar())==' '); if(c=='\n') break; la=c-'0'; } else la=la*10+c-'0'; } if(la) { nu[i]++; nu[la]++; ans++; } } //printf("ans:%d :%d %d\n",ans,nu[0],nu[1]); //system("pause"); int ocnt=0,a[3]; for(int i=0;i<n;i++) if(nu[i]&1) { ocnt++; if(ocnt>2) break; a[ocnt]=i; } gets(temp); if(ocnt>2||ocnt==1) printf("NO\n"); else if(!ocnt) { if(!m) printf("YES %d\n",ans); else printf("NO\n"); } else { if(a[1]>a[2]) swap(a[1],a[2]); if(!a[1]&&a[2]==m) printf("YES %d\n",ans); else printf("NO\n"); } } return 0; }
[歐拉回路] poj 1300 Door Man