poj 2549 Sumsets 折半枚举

http://poj.org/problem?id=2549

题意：给你n个数字，从中挑出四个数字使得a+b+c=d求d得最大值；

思想：折半枚举，否则n^4的复杂度；不过很多细节地方要注意；还要注意lower_bound在结构体排序的

应用；

#include <iostream>
#include<cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
struct Node{
int v, x, y, p, d;
bool operator <(const Node& b) const
{
return v<b.v;
}
};
Node nodeb[1000005], nodec[1000005];
int a[1005];
bool cmp(Node a, Node b)
{
return a.v<b.v;
}
int main()
{
int n;
while (~scanf("%d", &n)&&n)
{
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
int cnt1 = 0, cnt2 = 0, ans = 0, flag = 0;
for (int i = 1; i <= n-1; i++)
for (int j = i + 1; j <= n; j++)
{
++cnt1;
nodeb[cnt1].v = a[i] + a[j];
nodeb[cnt1].x =i;
nodeb[cnt1].y =j;
};
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (i == j)
continue;
else
{
++cnt2;
nodec[cnt2].v = a[i]- a[j];
nodec[cnt2].x = i;
nodec[cnt2].y = j;
nodec[cnt2].d = a[i];
};
sort(nodec + 1, nodec + cnt2 + 1, cmp);
for (int i = 1; i <= cnt1; i++)
{
int m = nodeb[i].v;
Node temp; temp.v = m;
int k = lower_bound(nodec + 1, nodec + cnt2 + 1, temp) - nodec;
if (m==nodec[k].v)
if (nodec[k].x != nodeb[i].x&&nodec[k].x != nodeb[i].y
&&nodec[k].y != nodeb[i].x&&nodec[k].y != nodeb[i].y)
{
flag = 1;
ans = ans<nodec[k].d ? nodec[k].d : ans;
}
}
if (!flag)
printf("no solution\n");
else
printf("%d\n", ans);
}
return 0;
}

下面是第一次wa代码：

#include <iostream>
#include<cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
struct Node{
   int v,x,y,p,d;
   bool operator <(const Node& b) const
   {
       return v<b.v;
   }
};
Node nodea[1005],nodeb[1005],nodec[1005];;
bool cmp(Node a,Node b)
{
    return a.v<b.v;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
            {
                scanf("%d",&nodea[i].v);
                nodea[i].p=i;
            }
        int cnt1=0,cnt2=0,ans=0,flag=0;
        for(int i=1;i<=n;i++)
            for(int j=i+1;j<=n;j++)
            if(i==j)
                continue;
            else
                {
                   nodeb[++cnt1].v=nodea[i].v+nodea[j].v;
                   nodeb[cnt1].x=nodea[i].p;
                   nodeb[cnt1].y=nodea[j].p;
                };
        for(int i=1;i<=n;i++)
            for(int j=i;j<=n;j++)
             if(i==j)
                continue;
            else
                {
                   nodec[++cnt2].v=nodea[i].v-nodea[j].v;
                   nodec[cnt2].x=nodea[i].p;
                   nodec[cnt2].y=nodea[j].p;
                   nodec[cnt2].d=nodea[i].v;
                };
        sort(nodec+1,nodec+cnt2+1,cmp);
        for(int i=1;i<=cnt1;i++)
            {
                int m=nodeb[i].v;
                Node temp;temp.v=-m;
                int k=lower_bound(nodec+1,nodec+cnt2+1,temp)-nodec;
                if(m+nodec[k].v==0)
                if(nodec[k].x!=nodeb[i].x&&nodec[k].x!=nodeb[i].y
                   &&nodec[k].y!=nodeb[i].x&&nodec[k].y!=nodeb[i].y)
                {
                    flag=1;
                    ans=ans<nodec[k].d?nodec[k].d:ans;
                }
            }
            if(!flag)
                printf("no solution\n");
            else
                printf("%d\n",ans);
    }
    return 0;
}