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题意:一个团队要去参观一些学校,某些学校要在某些学校之前先参观,并且每个学校有一个权值,团队去的时间与权值的差作为难过度(最小是0),
所有的难过度的最大值是伤心度,让你安排参观顺序,使得这个伤心度最小。
析:拓扑排序,并且要逆序排,这样的话,时间大的优先,可以用优先队列实现。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> //#include <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; //using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 5; const LL mod = 10000000000007; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } struct Node{ int id, val; Node() { } Node(int i, int v) : id(i), val(v) { } bool operator < (const Node& p) const{ return val < p.val; } }; vector<int> G[maxn]; int in[maxn], a[maxn]; int ans[maxn]; int main(){ freopen("grand.in", "r", stdin); freopen("grand.out", "w", stdout); while(scanf("%d", &n) == 1 && n){ for(int i = 1; i <= n; ++i){ scanf("%d", a+i); G[i].clear(); } scanf("%d", &m); int u, v; for(int i = 0; i < m; ++i){ scanf("%d %d", &u, &v); G[v].push_back(u); ++in[u]; } priority_queue<Node> pq; for(int i = 1; i <= n; ++i) if(!in[i]) pq.push(Node(i, a[i])); int cnt = n, num = 0; while(!pq.empty()){ Node u = pq.top(); pq.pop(); num = Max(num, Max(0, cnt-u.val)); ans[cnt--] = u.id; int x = u.id; for(int j = 0; j < G[x].size(); ++j){ int t = G[x][j]; --in[t]; if(!in[t]) pq.push(Node(t, a[t])); } } printf("%d\n", num); printf("%d", ans[1]); for(int i = 2; i <= n; ++i) printf(" %d", ans[i]); printf("\n"); } return 0; }
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原文地址:http://www.cnblogs.com/dwtfukgv/p/5925796.html