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[310][310],f[310];
int main()
{
    int n,m,i,j,q;
    int kk = 1;
    while(cin>>n>>m>>q)
    {
        if(n==0&&m==0&&q==0) break;
        int u,v,c;
        memset(f,0,sizeof(f));
        for(i = 0 ; i < n ; i++)
        {
            for(j = 0 ;j < n ; j++)
            w[i][j] = INF;
            w[i][i] = 0;
        }
        for(i = 1; i <= m ; i++)
        {
            cin>>u>>v>>c;
            w[u][v] = min(w[u][v],c);
        }
        printf("Case %d:
",kk++);
        while(q--)
        {
            int t;
            cin>>t;
            if(t==1)
            {
                int x,y;
                cin>>x>>y;
                if(!f[x]||!f[y])
                printf("City %d or %d is not available.
",x,y);
                else if(w[x][y]==INF)
                puts("No such path.");
                else
                printf("%d
",w[x][y]);
            }
            else
            {
                int x;
                cin>>x;
                if(f[x])
                {
                    printf("City %d is already recaptured.
",x);
                    continue;
                }
                f[x] = 1;
                for(i = 0 ; i < n ; i++)
                    for(j = 0 ; j < n ; j++)
                    if(w[i][j]>w[i][x]+w[x][j])
                    w[i][j] = w[i][x]+w[x][j];
            }
        }
        puts("");
    }
    return 0;
}
 



/**************************************
    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[10];
int main()
{
    int n,kk=1;
    while(cin>>n)
    {
        if(!n) break;
        set<node>q;
        node st;
        printf("Case %d:
", kk++);
        while(n--)
        {
            cin>>s;
            if(s[0]=='a')
            {
                cin>>st.x>>st.y;
                q.insert(st);
            }
            else if(s[0]=='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 0;
}
 



/**************************************
    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[1010],p[1010*1010];
int main()
{
    int n,m,i,j;
    int kk = 1;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(!n&&!m) break;
        for(i = 0; i < n ; i++)
        scanf("%d",&a[i]);
        a[n] = 0;
        int g = 0;
        for(i = 0 ; 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 = 0;
        printf("Case %d: ",kk++);
        for(i = 0 ; i < g ;  i++)
        {
            if(p[i]>m) break;
            int low = i,high = g-1;
            while(low<high)
            {
                int mm = (low+high)/2;
                if(p[i]+p[mm]>m)
                high = mm-1;
                else
                low = mm+1;
            }
            if(p[low]+p[i]<=m&&p[low]+p[i]>maxz)
            {
                maxz = p[low]+p[i];
            }
        }
        printf("%d
",maxz);
        puts("");
    }
    return 0;
}
 



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