[PA 2014]Kuglarz

Description

魔术师的桌子上有n个杯子排成一行，编号为1,2,…,n，其中某些杯子底下藏有一个小球，如果你准确地猜出是哪些杯子，你就可以获得奖品。花费c_ij元，魔术师就会告诉你杯子i,i+1,…,j底下藏有球的总数的奇偶性。
采取最优的询问策略，你至少需要花费多少元，才能保证猜出哪些杯子底下藏着球？

Input

第一行一个整数n(1<=n<=2000)。
第i+1行(1<=i<=n)有n+1-i个整数，表示每一种询问所需的花费。其中c_ij（对区间[i,j]进行询问的费用，1<=i<=j<=n,1<=c_ij<=10^9）为第i+1行第j+1-i个数。

Output

输出一个整数，表示最少花费。

Sample Input

5
1 2 3 4 5
4 3 2 1
3 4 5
2 1
5

Sample Output

7

题解

求一棵最小生成树...

 //It is made by Awson on 2017.10.15
 #include <set>
 #include <map>
 #include <cmath>
 #include <ctime>
 #include <cmath>
 #include <stack>
 #include <queue>
 #include <vector>
 #include <string>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #define LL long long
 #define Min(a, b) ((a) < (b) ? (a) : (b))
 #define Max(a, b) ((a) > (b) ? (a) : (b))
 #define sqr(x) ((x)*(x))
 using namespace std;
 const int N = ;
 const int INF = ~0u>>;
 
 int n, mp[N+][N+];
 int dist[N+];
 bool vis[N+];
 
 LL Prim() {
     LL ans = ;
     for (int i = ; i <= n; i++) dist[i] = mp[][i];
     vis[] = ;
     for (int t = ; t < n; t++) {
         int loc, minn = INF;
         for (int i = ; i <= n; i++) if (!vis[i] && dist[i] < minn) {
             minn = dist[i], loc = i;
         }
         ans += minn; vis[loc] = ;
         for (int i = ; i <= n; i++) dist[i] = Min(dist[i], mp[loc][i]);
     }
     return ans;
 }
 void work() {
     scanf("%d", &n); n++;
     for (int i = ; i < n; i++) for (int j = i+; j <= n; j++) scanf("%d", &mp[i][j]), mp[j][i] = mp[i][j];
     printf("%lld\n", Prim());
 }
 int main() {
     work();
     return ;
 }