Airports
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DAG
考虑一架飞机能完成任务 i 之后能赶去完成任务 j,那么就在这两个任务之间连一条线,最后求的就是最少链覆盖所有点,即DAG。实际上就是把每个点拆成 x,y, 上述就是把
注意一个坑点就是这道题很显然要用floyd进行路径距离优化来判断飞机是否能在完成任务 i 之后去完成任务 j,但是原题要求是飞机飞行一定要是直达,即每个任务的飞行时间不能用经过Floyd优化后的邻接表来计算,被这英文坑了很久。
#include <bits/stdc++.h>#include <vector>using namespace std;const int N = 5000 + 10;bool dis[N][N];bool vis[N];int cy[N], a[N][N], p[N], n, m, ans, uN, vN, a1[N][N];void floyd() { for(int k = 1; k <= m; k++) for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j) if(dis[i][k] && dis[k][j])//传递可达性 dis[i][j] = true;}int linker[N];bool used[N];bool dfs(int u) { for(int v = 1; v <= vN; v++) if(dis[u][v] && !used[v]) { used[v] = true; if(linker[v] == -1 || dfs(linker[v])) { linker[v] = u; return true; } } return false; }int hungary() { uN = vN = m; int res = 0; memset(linker, -1, sizeof(linker)); for(int u = 1; u <= uN; u++) { memset(used, false, sizeof(used)); if (dfs(u)) res++; } return res;}struct Node { int s, f, t; Node() {} bool operator<(const Node other) const { return t<other.t; }};Node b[N];int main() { scanf("%d %d", &n, &m); for (int i = 1; i <= n; i++) scanf("%d", &p[i]); for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) { scanf("%d", &a[i][j]); if (i != j) a[i][j] += p[j]; a1[i][j] = a[i][j]; } for (int k = 1; k <= n; k++) for (int i = 1; i <= n; i++) if (i != k) for (int j = 1; j <= n; j++) if (i != j && k != j) { a[i][j] = min(a[i][j], a[i][k]+a[k][j]); } for (int i = 1; i <= m; i++) for (int j = 1; j <= m; j++) dis[i][j] = false; for (int i = 1; i <= m; i++) scanf("%d %d %d", &b[i].s, &b[i].f, &b[i].t); sort(b+1, b+1+m); for (int i = 1; i <= m; i++) for (int j = i+1; j <= m; j++) if (b[i].t+a1[b[i].s][b[i].f]+a[b[i].f][b[j].s] <= b[j].t) { dis[i][j] = true; } ans = m-hungary(); printf("%d\n", ans); return 0;}
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