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分析

貪心地想，對于兩座城市之間，有多條道路，為了每次能多運輸一些貨物，一定會盡量往邊權較大的路走，所以，一些邊權較小的路是不會被走過，于是考慮 \(Kruskal\) 重構樹，然后在新的圖上跑 \(LCA\) 求出路徑中邊權最小的值。

代碼

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 10003
#define M 50003
#define il inline
#define re register
#define INF 0x3f3f3f3f
#define tie0 cin.tie(0),cout.tie(0)
#define fastio ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;template <typename T> inline void read(T &x) {T f = 1; x = 0; char c;for (c = getchar(); !isdigit(c); c = getchar()) if (c == '-') f = -1;for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);x *= f;
}struct edge1 {int u, v, w;
} e1[M];struct edge2 {int to, nxt, val;
} e2[M];int n, m, q;
int fa[N][25], dep[N], f[N], dis[N][25];
int head[M], cnt;
bool vis[N];bool cmp(edge1 x, edge1 y) {return x.w > y.w;
}void prefix() {for (int j = 1; j <= 20; ++j)for (int i = 1; i <= n; ++i) {fa[i][j] = fa[fa[i][j-1]][j-1];dis[i][j] = min(dis[fa[i][j-1]][j-1], dis[i][j-1]);}
}int find(int x) {return f[x] == x ? x : f[x] = find(f[x]);
}void addedge(int u, int v, int w) {e2[++cnt].to = v, e2[cnt].val = w, e2[cnt].nxt = head[u], head[u] = cnt;e2[++cnt].to = u, e2[cnt].val = w, e2[cnt].nxt = head[v], head[v] = cnt;
}void Kruskal() {for (int i = 1; i <= m; ++i) {int f1 = find(e1[i].u), f2 = find(e1[i].v);if (f1 == f2) continue;else {addedge(e1[i].u, e1[i].v, e1[i].w);f[f1] = f2;}}
}void dfs(int u) {vis[u] = 1;for (int i = head[u]; i; i = e2[i].nxt) {int v = e2[i].to;if (vis[v]) continue;fa[v][0] = u, dis[v][0] = e2[i].val;dep[v] = dep[u] + 1;dfs(v);}
}int LCA(int x, int y) {int ans = INF;if (find(x) != find(y)) return -1;if (dep[x] < dep[y]) swap(x, y);for (int i = 20; i >= 0; --i)if (dep[fa[x][i]] >= dep[y]) {ans = min(ans, dis[x][i]);x = fa[x][i];}if (x == y) return ans;for (int i = 20; i >= 0; --i)if (fa[x][i] != fa[y][i]) {ans = min(ans, min(dis[x][i], dis[y][i]));x = fa[x][i], y = fa[y][i];}return min(ans, min(dis[x][0], dis[y][0]));
}int main() {int x, y;read(n), read(m);for (int i = 1; i <= n; ++i) f[i] = i;for (int i = 1; i <= m; ++i) read(e1[i].u), read(e1[i].v), read(e1[i].w);sort(e1 + 1, e1 + 1 + m, cmp);Kruskal();for (int i = 1; i <= n; ++i)if (!vis[i]) {dep[i] = 1;dfs(i);fa[i][0] = i, dis[i][0] = INF;}prefix();read(q);while (q--) {read(x), read(y);printf("%d\n", LCA(x, y));}return 0;
}

轉載于:https://www.cnblogs.com/hlw1/p/11437340.html

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