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Time Limit: 20 Sec Memory Limit: 256 MB
http://acm.hdu.edu.cn/showproblem.php?pid=5232
Multiple test cases (about 30), the first line of each case contains an integer n which indicates Gorwin will invite n friends to her party.
Next
n lines will give a n*n matrix, if a[i][j] is 1, then friend i and
friend j have known each other, otherwise they have not known each
other.
Please process to the end of file.
[Technical Specification]
All input entries are integers.
1<=n<=30
0<=a[i][j]<=1
a[i][i]=0;
a[i][j]=a[j][i] for i!=j
Output
For each case, output an integer which denotes total cups of champagne Gorwin should prepare in a single line.
4 8
For the second case, Gorwin will shake hands with all her friends, then Gorwin drink three cups of champagne, each friends drink one cup. Friend 1 and friend 3 know each other,every of them drinks one cup again. So the total cups is 3+3+2=8.
题意
今天是Gorwin的生日，所以她举办了一个派对并邀请她的朋友来参加。她将邀请n个朋友，为了方便，Gorwin把他们从1到n标号。他们之中有一些人已经相互认识，有一些人不认识对方。相互认识的朋友见面之后会握手然后喝一杯香槟。Gorwin想要知道要准备多少杯香槟。你能帮助她吗？
题解：
给你的图里面有多少个1,ans就加多少
然后再加n，再乘2就好了
代码：
//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 200001 #define mod 10007 #define eps 1e-9 int Num; char CH[20]; //const int inf=0x7fffffff; //нчоч╢С const int inf=0x3f3f3f3f; /* inline void P(int x) { Num=0;if(!x){putchar(‘0‘);puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts(""); } */ inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } inline void P(int x) { Num=0;if(!x){putchar(‘0‘);puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts(""); } //************************************************************************************** int g[300][300]; int main() { //freopen("test.txt","r",stdin); int n; while(scanf("%d",&n)!=EOF) { memset(g,0,sizeof(g)); for(int i=0;i<n;i++) for(int j=0;j<n;j++) g[i][j]=read(); int ans=0; for(int i=0;i<n;i++) { for(int j=0;j<n;j++) { if(g[i][j]) { ans++; g[i][j]=0; g[j][i]=0; } } } cout<<(ans+n)*2<<endl; } }
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原文地址：http://www.cnblogs.com/qscqesze/p/4524900.html
