2010山东省第一届ACM程序设计竞赛

休眠了2月了 要振作起来了！！。。。

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2155

因为点比较少 最多更新三百次 标记某个节点时直接更新与之相连的点的最短距离

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 using namespace std;
 #define INF 0xfffffff
 int w[][],f[];
 int main()
 {
     int n,m,i,j,q;
     int kk = ;
     while(cin>>n>>m>>q)
     {
         if(n==&&m==&&q==) break;
         int u,v,c;
         memset(f,,sizeof(f));
         for(i =  ; i < n ; i++)
         {
             for(j =  ;j < n ; j++)
             w[i][j] = INF;
             w[i][i] = ;
         }
         for(i = ; i <= m ; i++)
         {
             cin>>u>>v>>c;
             w[u][v] = min(w[u][v],c);
         }
         printf("Case %d:\n",kk++);
         while(q--)
         {
             int t;
             cin>>t;
             if(t==)
             {
                 int x,y;
                 cin>>x>>y;
                 if(!f[x]||!f[y])
                 printf("City %d or %d is not available.\n",x,y);
                 else if(w[x][y]==INF)
                 puts("No such path.");
                 else
                 printf("%d\n",w[x][y]);
             }
             else
             {
                 int x;
                 cin>>x;
                 if(f[x])
                 {
                     printf("City %d is already recaptured.\n",x);
                     continue;
                 }
                 f[x] = ;
                 for(i =  ; i < n ; i++)
                     for(j =  ; j < n ; j++)
                     if(w[i][j]>w[i][x]+w[x][j])
                     w[i][j] = w[i][x]+w[x][j];
             }
         }
         puts("");
     }
     return ;
 }

 /**************************************
     Problem id    : SDUT OJ 2155
     User name    : shang
     Result        : Accepted
     Take Memory    : 832K
     Take Time    : 300MS
     Submit Time    : 2014-01-16 15:11:20
 **************************************/

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2159

set存结构体 low_bounder二分查找 不过时间跑了不少 不知道有没有更简单的方法

 #include<cstdio>
 #include<cstring>
 #include <cmath>
 #include<algorithm>
 #include<stdlib.h>
 #include<queue>
 #include<vector>
 #include<set>
 #include<iostream>
 using namespace std;
 struct node
 {
     int x,y;
     bool operator <(const node a) const
     {
         if(a.x==x)
         return y<a.y;
         return x<a.x;
     }
 };
 char s[];
 int main()
 {
     int n,kk=;
     while(cin>>n)
     {
         if(!n) break;
         set<node>q;
         node st;
         printf("Case %d:\n", kk++);
         while(n--)
         {
             cin>>s;
             if(s[]=='a')
             {
                 cin>>st.x>>st.y;
                 q.insert(st);
             }
             else if(s[]=='f')
             {
                 cin>>st.x>>st.y;
                 set<node>::iterator it;
                 it = q.lower_bound(st);
                 while(it!=q.end())
                 {
                     if((*it).x>st.x&&(*it).y>st.y)
                     {
                         cout<<(*it).x<<" "<<(*it).y<<endl;
                         break;
                     }
                     it++;
                 }
                 if(it==q.end())
                 puts("-1");
             }
             else
             {
                 cin>>st.x>>st.y;
                 q.erase(st);
             }
         }
         puts("");
     }
     return ;
 }

 /**************************************
     Problem id    : SDUT OJ 2159
     User name    : shang
     Result        : Accepted
     Take Memory    : 2572K
     Take Time    : 930MS
     Submit Time    : 2014-01-16 17:34:20
 **************************************/

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2157

先取两个数 再二分查找和与m最近的数

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 using namespace std;
 int a[],p[*];
 int main()
 {
     int n,m,i,j;
     int kk = ;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         if(!n&&!m) break;
         for(i = ; i < n ; i++)
         scanf("%d",&a[i]);
         a[n] = ;
         int g = ;
         for(i =  ; i <= n; i++)
             for(j = i ; j <= n ; j++)
             if(a[i]+a[j]<=m)
             p[g++] = a[i]+a[j];
         sort(p,p+g);
         int maxz = ;
         printf("Case %d: ",kk++);
         for(i =  ; i < g ;  i++)
         {
             if(p[i]>m) break;
             int low = i,high = g-;
             while(low<high)
             {
                 int mm = (low+high)/;
                 if(p[i]+p[mm]>m)
                 high = mm-;
                 else
                 low = mm+;
             }
             if(p[low]+p[i]<=m&&p[low]+p[i]>maxz)
             {
                 maxz = p[low]+p[i];
             }
         }
         printf("%d\n",maxz);
         puts("");
     }
     return ;
 }

 /**************************************
     Problem id    : SDUT OJ 2157
     User name    : shang
     Result        : Accepted
     Take Memory    : 2444K
     Take Time    : 130MS
     Submit Time    : 2014-01-16 15:43:45
 **************************************/