CodeForces 358E - Dima and Kicks

来源：互联网发布：九维外呼软件编辑：程序博客网时间：2024/05/21 09:52

dfs判断欧拉图，红名选手的代码就是炫酷。

首先统计所有点的度数总和，而后对于这张图的特殊性——每个点最多只会有四条边，来标记当前边是否走过了。

若在一次DFS中，能遍历所有的节点则输出所有边长的gcd的大于1的约数集。

真心学习了。

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#include <cmath>#include <stack>#include <map>#pragma comment(linker, "/STACK:1024000000");#define EPS (1e-8)#define LL long long#define ULL unsigned __int64#define _LL __int64#define INF 0x3f3f3f3f#define Mod 20090717using namespace std;bool vis[1002][1002][4];int num[1010][1010];int dir[2][4] = {{-1,0,1,0},{0,-1,0,1}};int total;int ans;void dfs(int x,int y,int d,int len){    bool mark = false;    for(int i = 0;i < 4; ++i)    {        if(0 == num[x+dir[0][i]][y+dir[1][i]] || vis[x][y][i] == true)            continue;        vis[x][y][i] = true,vis[x+dir[0][i]][y+dir[1][i]][(i+2)%4] = true;        total -= 2;        if(d != -1 && i != d)            ans = __gcd(ans,len),dfs(x+dir[0][i],y+dir[1][i],i,1);        else            dfs(x+dir[0][i],y+dir[1][i],i,len+1);         mark = true;    }    if(mark == false)        ans = __gcd(ans,len);}int main(){    int n,m,i,j,k;    scanf("%d %d",&n,&m);    memset(num,0,sizeof(num));    for(i = 1;i <= n; ++i)        for(j = 1;j <= m; ++j)            scanf("%d",&num[i][j]);    int odd = 0;    ans = 0;    total = 0;    for(i = 1;i <= n; ++i)        for(j = 1;j <= m; ++j)        {            if(num[i][j] == 0)                continue;            for(ans = 0,k = 0;k < 4; ++k)                ans += num[i+dir[0][k]][j+dir[1][k]];            total += ans;            if(ans == 0)                return puts("-1"),0;            if(ans&1)                odd++;        }    if(odd != 0 && odd != 2)        return puts("-1"),0;    memset(vis,false,sizeof(vis));    ans = 0;    for(i = 1;i <= n; ++i)        for(j = 1;j <= m; ++j)            if(num[i][j])            {                dfs(i,j,-1,0);                goto V;            }    V:;    if(total || ans <= 1)        return puts("-1"),0;    for(i = 2;i < ans; ++i)        if(ans%i == 0)            printf("%d ",i);    printf("%d\n",ans);    return 0;}

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