【Codeforces Round 323 (Div 2)A】【水题】Asphalting Roads 行列之进行首次操作
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#include<stdio.h>#include<string.h>#include<ctype.h>#include<math.h>#include<iostream>#include<string>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}const int N=55,M=0,Z=1e9+7,ms63=1061109567;int casenum,casei;int n,y,x;int line[N],list[N];int main(){while(~scanf("%d",&n)){MS(line,0);MS(list,0);for(int i=1;i<=n*n;i++){scanf("%d%d",&y,&x);if(!line[y]&&!list[x]){printf("%d ",i);line[y]=list[x]=1;}}puts("");}return 0;}/*【题意】给你n*n(n∈[1,50])个操作,每个操作选取一个二维平面上的点(横纵坐标都是[1,n]范围)。如果对于一个操作,其横纵坐标上之前都没有选过点,那么我们会选择这个操作执行;否则这个操作不执行。现在让你输出所有执行的操纵。【类型】水题【分析】直接暴力哈希行列。*/
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