鏈接：https://ac.nowcoder.com/acm/contest/882/F
來源：牛客網
Given 2N people, you need to assign each of them into either red team or white team such that each team consists of exactly N people and the total competitive value is maximized.
Total competitive value is the summation of competitive value of each pair of people in different team.
The equivalent equation is
∑2Ni=1∑2Nj=i+1(vij if i-th person is not in the same team as j-th person else 0)∑i=12N∑j=i+12N(vij if i-th person is not in the same team as j-th person else 0)
輸入描述:
The first line of input contains an integers N.
Following 2N lines each contains 2N space-separated integers vijvij is the j-th value of the i-th line which indicates the competitive value of person i and person j.
* 1≤N≤141≤N≤14
* 0≤vij≤1090≤vij≤109
* vij=vjivij=vji
輸出描述:
Output one line containing an integer representing the maximum possible total competitive value.
示例1
輸入
1 0 3 3 0
輸出
3
題意：
將n*2個人分為兩部分，每一個人與另外一半的每一個人貢獻一個權值，求貢獻和的最大值。
思路：
暴力搜索，最壞的復雜度是C(2*14,14)，也就是差不多4e7,如果你確定某一個人在第一部分，還可以將復雜度除而2
關於算貢獻，你可以選擇14*14的復雜度，但是肯定會T
在搜索的時候，如果n=5,那么第一次選在第一部分的人就是 1 2 3 4 5.
第二次選在第一部分的人就是 1 2 3 4 6,可以發現只有一個數字不同。
分析一下，其實在整個搜索的過程中，也會出現很多這樣只差一個的數組。
於是，我們可以記錄上一個狀態，通過上個狀態算出當前狀態，這樣可以減小很多算貢獻的復雜度。
就這樣，我的代碼跑了3700ms之后卡過去了。
#include <bits/stdc++.h> #define eps 1e-8 #define INF 0x3f3f3f3f #define PI acos(-1) #define lson l,mid,rt<<1 #define rson mid+1,r,(rt<<1)+1 #define CLR(x,y) memset((x),y,sizeof(x)) #define fuck(x) cerr << #x << "=" << x << endl; using namespace std; typedef long long ll; typedef unsigned long long ull; const int seed = 131; const int maxn = 1e5 + 5; const int mod = 1e9 + 7; int n; ll mp[30][30]; bool vis[30]; ll MIN; int v1[30]; int v2[30]; int prev1[30]; int prev2[30]; ll prenum = 0; ll C[35][35]; //C[n][m]就是C(n,m) int tot; void init(int N) { for (int i = 0; i < N; i ++) { C[i][0] = C[i][i] = 1; for (int j = 1; j < i; j ++) { C[i][j] = C[i - 1][j] + C[i - 1][j - 1]; C[i][j] %= mod; } } } void dfs(int x, int deep) {//必須>=x開始,已經選了num個人 if (deep == n) { tot--; if(tot<0){return;} int cnt1 = 0; int cnt2 = 0; for (int i = 1; i <= 2 * n; i++) { if (vis[i]) v1[++cnt1] = i; else v2[++cnt2] = i; } ll num = prenum; int pos = 1; for (int i = 1; i <= n; i++) { if (v1[i] != prev1[i]) { pos = i; break; } } for (int i = pos; i <= n; i++) { for (int j = 1; j <= n; j++) { num -= mp[prev1[i]][prev2[j]]; num -= mp[v1[i]][prev1[j]]; } for (int j = 1; j <= n; j++) { num += mp[v1[i]][v2[j]]; num += mp[v1[j]][prev1[i]]; } } MIN = max(MIN, num); for (int i = 1; i <= n; i++) { prev1[i] = v1[i]; prev2[i] = v2[i]; prenum = num; } return ; } for (int i = x + 1; i <= 2 * n; i++) { vis[i] = 1; dfs(i, deep + 1); if(tot<0){return;} vis[i] = 0; } } int main() { MIN = -1; init(30); scanf("%d", &n); tot=C[2*n][n]; tot/=2; for (int i = 1; i <= 2 * n; i++) { for (int j = 1; j <= 2 * n; j++) { scanf("%lld", &mp[i][j]); } } dfs(0, 0); printf("%lld\n", MIN); return 0; }