uva_10125_Sumsets( 二分 )

題意：

給你一個整數集合，要求你從中找出一個最大的d，使得a+b+c=d,其中a,b,c,d都是集合s中的不同元素

分析：

如果暴力的話O(N^4），但是可以使用二分解決，就是美劇a+b後對其結果排序，接着枚舉d-c如果二分查找a+b的結果中存在a+b = d-c，那麼這個d就是一個解

時間複雜度O(N^2*log(N*N))

Code:

#include <set>#include <map>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <deque>#include <vector>#include <cstdio>#include <bitset>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;#define DIR     4#define DIM     2#define STATUS  2#define MAXN    1000000 + 10#define oo      (~0u)>>1#define INF     0x3F3F3F3F#define REPI(i, s, e)   for(int i = s; i <= e; i ++)#define REPD(i, e, s)   for(int i = e; i >= s; i --)static const double EPS = 1e-5;typedef struct Node_ {        long long sum, each;}Node;Node s[MAXN];long long val[MAXN], arr[MAXN];void input(int n){        REPI(i, 1, n) {                scanf("%lld", &val[i]);        }}int valid(long long a, long long b, long long c, long long d){        if( a == b || a == c || a == d ) {                return 0;        }        if( b == c || b == d || c == d ) {                return 0;        }        return 1;}inline int cmp(const Node &a, const Node &b){        return a.sum < b.sum;}long long process(int n){        int idx = 0;        REPI(i, 1, n) {                REPI(j, i+1, n) {                        if( val[i] == val[j] ) {                                continue;                        }                        s[idx].each = val[i];                        s[idx ++].sum = val[i]+val[j];                }        }        sort(s, s+idx, cmp);        REPI(i, 0, idx-1) {                arr[i] = s[i].sum;        }        long long rst = -INF;        REPI(i, 1, n) {                REPI(j, 1, n) {                        long long max_one = val[i], min_one = val[j];                        if( max_one < min_one ) {                                swap(max_one, min_one);                        }                        else if( max_one == min_one ) {                                continue;                        }                        int m = lower_bound(arr, arr+idx, max_one-min_one)-arr;                        if( arr[m] != max_one-min_one ) {                                continue;                        }                        if( !valid(max_one, min_one, s[m].each, s[m].sum-s[m].each) ) {                                continue;                        }                        rst = max(rst, val[i]);                }        }        return rst;}void ouput(long long rst){        if( -INF == rst ) {                printf("no solution\n");                return;        }        printf("%lld\n", rst);}int main(int argc, char const *argv[]){#ifndef ONLINE_JUDGE        freopen("test.in", "r", stdin);#endif        int n;        while( scanf("%d", &n) && n ) {                input(n);                ouput(process(n));        }        return 0;}