poj 3253 优先队列

Fence RepairTime Limit: 2000MS      Memory Limit: 65536KTotal Submissions: 47005        Accepted: 15345DescriptionFarmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length Li (1 ≤ Li ≤ 50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths Li). FJ is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made; you should ignore it, too.FJ sadly realizes that he doesn't own a saw with which to cut the wood, so he mosies over to Farmer Don's Farm with this long board and politely asks if he may borrow a saw.Farmer Don, a closet capitalist, doesn't lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will result in different charges since the resulting intermediate planks are of different lengths.InputLine 1: One integer N, the number of planks Lines 2..N+1: Each line contains a single integer describing the length of a needed plankOutputLine 1: One integer: the minimum amount of money he must spend to make N-1 cutsSample Input3858Sample Output34HintHe wants to cut a board of length 21 into pieces of lengths 8, 5, and 8. The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into 16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).SourceUSACO 2006 November Gold

分析:每次取最小的两个数,合成一个后再放回去,直到得到最终结果.

//priority_queue<int,vector<int,greater<int>> qu;//第一个int表示数据类型,第二个vector表示存储数据的容器,第三个表示自定义的优先级规则,greater(从小到大),或者less(从大到小)都行#include <cstdio>#include <cstdlib>#include <iostream>#include <stack>#include <queue>#include <algorithm>#include <cstring>#include <cmath>#include <vector>#include <bitset>#include <list>#include <sstream>#include <set>#include <functional>using namespace std;#define INT_MAX 1 << 30#define MAX 100typedef long long ll;int main(int argc, char const* argv[]){    //input    int n;    int in;    priority_queue<int,vector<int>,greater<int>> qu;    scanf("%d",&n);    for (int i = 0; i < n; i += 1){        scanf("%d",&in);        qu.push(in);    }    //solve    ll ans = 0;    while (qu.size() > 1){        int a,b;        a = qu.top();        qu.pop();        b = qu.top();        qu.pop();        qu.push(a+b);        ans += a+b;    }    printf("%lld\n",ans);    return 0;}