标签:type string family inf ace unsigned cst int lis
题意:有n(n <= 5000)个数的集合S,每次可以从S中删除两个数,然后把它们的和放回集合,直到剩下一个数。每次操作的开销等于删除的两个数之和,求最小总开销。所有数均小于10^5。
分析:按此操作,最终变成1个数,需要n-1次操作,要想总开销最小,就使每次取出的两数之和最小,优先队列。
#pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) typedef long long LL; typedef unsigned long long ULL; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const LL LL_INF = 0x3f3f3f3f3f3f3f3f; const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const double eps = 1e-8; const int MAXN = 500 + 10; const int MAXT = 10000 + 10; using namespace std; priority_queue<int, vector<int>, greater<int> > q; int main(){ int n; while(scanf("%d", &n) == 1){ if(!n) return 0; while(!q.empty()) q.pop(); for(int i = 0; i < n; ++i){ int x; scanf("%d", &x); q.push(x); } int ans = 0; while(q.size() > 1){ int a = q.top(); q.pop(); int b = q.top(); q.pop(); ans += a + b; q.push(a + b); } printf("%d\n", ans); } return 0; }
UVA - 10954 Add All (全部相加)(Huffman编码 + 优先队列)
标签:type string family inf ace unsigned cst int lis
原文地址:http://www.cnblogs.com/tyty-Somnuspoppy/p/6371404.html