POJ - 2431 Expedition(贪心+优先队列)
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题意:已知某车距离城镇为L,油量为P,油箱无穷大,已知有n个加油量为x的加油站,问,车到城镇最少加几次油。若不能到达城镇,则输出-1。
分析:
1、贪心,先将加油站按照离城镇由远及近排序。
2、卡车只要油够,就不断往前走,若当前油量不足以到达终点(或下一个加油站),则在之前经过的加油站里选择提供油量最大的加油站,加够了再继续走。
#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) const double eps = 1e-8; inline int dcmp(double a, double b){ if(fabs(a - b) < eps) return 0; return a > b ? 1 : -1; } 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 int MAXN = 10000 + 10; const int MAXT = 10000 + 10; using namespace std; int N, L, P; struct Node{ int d, fuel; Node(){} void read(){ scanf("%d%d", &d, &fuel); } void Set(int dd, int ff){ d = dd; fuel = ff; } bool operator < (const Node& rhs)const{ return d > rhs.d; } }num[MAXN]; priority_queue<int> q; int solve(){ int ans = 0; for(int i = 1; i <= N + 1; ++i){ int tmp = num[i - 1].d - num[i].d; if(P - tmp >= 0) P -= tmp; else{ while(!q.empty() && P - tmp < 0){ P += q.top(); q.pop(); ++ans; } if(P - tmp < 0) return -1; P -= tmp; } q.push(num[i].fuel); } return ans; } int main(){ scanf("%d", &N); for(int i = 1; i <= N; ++i){ num[i].read(); } scanf("%d%d", &L, &P); sort(num + 1, num + 1 + N); num[0].Set(L, 0); num[N + 1].Set(0, 0); printf("%d\n", solve()); return 0; }
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