题意：有一个人有n头猪，他在从A镇到B镇的n个村庄中每个村庄卖一头猪，每个村庄猪的价格不同，有运费，问他最多能卖多少钱、
分析：很容易想到猪运的越远所花路费越多，因此每个村庄猪的实际售价实质等于猪的收购价-从A镇运到该镇的路费。
首先算出每个村庄的实际售价，剩下的就是匹配问题，当然猪的重量与村庄的实际售价成正比赚钱越多。
#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 llu; 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}; const int dc[] = {-1, 1, 0, 0}; const double pi = acos(-1.0); const double eps = 1e-8; const int MOD = 1e9 + 7; const int MAXN = 1000 + 10; const int MAXT = 10000 + 10; using namespace std; int ans[MAXN]; struct P1 { int id; ll w; bool operator < (const P1 & a)const { return w > a.w; } } num1[MAXN]; struct P2 { int id; ll d, p, real; bool operator < (const P2 & a)const { return real > a.real; } } num2[MAXN]; int main() { int n; ll t; while(scanf("%lld%lld", &n, &t) == 2) { for(int i = 0; i < n; ++i) { scanf("%lld", &num1[i].w); num1[i].id = i + 1; } for(int i = 0; i < n; ++i) scanf("%lld", &num2[i].d); for(int i = 0; i < n; ++i) { num2[i].id = i + 1; scanf("%lld", &num2[i].p); num2[i].real = num2[i].p - num2[i].d * t; } sort(num1, num1 + n); sort(num2, num2 + n); for(int i = 0; i < n; ++i) { ans[num2[i].id] = num1[i].id; } for(int i = 1; i <= n; ++i) { if(i > 1) printf(" "); printf("%d", ans[i]); } printf("\n"); } return 0; }