Gym 100512G Grand Tour (拓扑排序)

Posted 2020-08-12 dwtfukgv

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;
}