Light OJ 1049 - One Way Roads （暴力 or DFS）

传送门：http://lightoj.com/volume_showproblem.php?problem=1049

题目大意：给定n个地点，n个点连成一个环，点与点之间是有向边，求使得图成为强连通图的最小花费。

解题思路：n范围比较小，两层FOR暴力或者DFS找出正向边和反向边即可，然后比较两个的权值和大小，即可求出答案。

Code：

/*   W          w           w        mm          mm             222222222       7777777777777    *//*    W        w w         w        m  m        m  m          222        22              7777    *//*    w        w w         w        m  m        m  m                     22              777     *//*     w      w   w       w        m    m      m    m                    22              77      *//*     w      w    w      w        m    m      m    m                 222                77      *//*      w    w      w    w        m      m    m      m              222                  77      *//*      w    w      w    w        m      m    m      m            222                    77      *//*       w  w        w  w        m        m  m        m         222                      77      *//*       w  w        w  w        m        m  m        m      222                         77      *//*        ww          ww        m          mm          m     222222222222222             77      *///#pragma comment(linker, "/STACK:102400000,102400000")//C++//int size = 256 << 20; // 256MB//char *p = (char*)malloc(size) + size;//__asm__("movl %0, %%esp\n" :: "r"(p));//G++#include<set>#include<map>#include<queue>#include<stack>#include<ctime>#include<deque>#include<cmath>#include<vector>#include<string>#include<cctype>#include<cstdio>#include<cstdlib>#include<cstring>#include<sstream>#include<iostream>#include<algorithm>#define REP(i,s,t) for(int i=(s);i<=(t);i++)#define REP2(i,t,s) for(int i=(t);i>=s;i--)using namespace std;typedef long long ll;typedef unsigned long long ull;typedef unsigned long ul;const int N=105;int T;int n;struct Edge{    int u,v,w;}E[N];bool vis[N];int main(){  #ifdef ONLINE_JUDGE  #else    freopen("test.in","r",stdin);  #endif  int ca=1;  scanf("%d",&T);  while(T--)  {    scanf("%d",&n);    int tot=0;    REP(i,1,n)    {        scanf("%d%d%d",&E[i].u,&E[i].v,&E[i].w);        tot+=E[i].w;    }    int res=0;    memset(vis,0,sizeof(vis));    int cnt=1;    int curu=E[1].u,curv=E[1].v;    vis[1]=1;    while(1)    {        if(cnt==n)        {            break;        }        REP(i,1,n)        {            if(!vis[i])            {                if(curv==E[i].u)                {                    vis[i]=1;                    curu=E[i].u;                    curv=E[i].v;                    cnt++;                    break;                }                else if(curv==E[i].v)                {                    vis[i]=1;                    curu=E[i].v;                    curv=E[i].u;                    res+=E[i].w;                    cnt++;                    break;                }            }        }    }    printf("Case %d: %d\n",ca++,min(res,tot-res));  }  return 0;}