題目鏈接

題意

有$A、B、C$三個人，$A$會在集合$Sa$中等概率出現，$B$會在結合$Sb$中等概率出現，$C$會等概率出現在所有點上。求他們三個人回合的最小路徑和的期望

思路

• 預處理$Sa、Sb$兩個集合到其他點的距離（BFS）
• 枚舉A、B的出現的所有狀態，對每個點初始$dis[i]=D[A][i]+D[B][i]$，對所有點跑最短路更新$dis$，求得每個點作為C出現對答案的貢獻。

因為所有的邊權都相同，可以用兩個隊列維護將最短路優化的O(n)，另外加上基數排序，可以卡過時間

枚舉狀態，得到每種狀態最長的回合距離，然后重新建圖，距離遞增建邊，每個距離對應的點建邊，最后跑BFS就行（邊權相同處理），因為每個點的的距離都增加1，所以最后要減去

// 做法1 #include <bits/stdc++.h> const int maxn = 1e5 + 5; const int inf = 0x3f3f3f3f; using namespace std; vector<int> g[maxn]; int D[2][30][maxn], a[30], b[30], dis[maxn]; int num[maxn<<1], Rank[maxn]; int n, m;void bfs(int s, int *d) { for (int i = 1; i <= n; ++i) {d[i] = -1;}d[s] = 0;queue<int> que;que.push(s);while (!que.empty()) {int u = que.front();que.pop();for (int i = 0; i < (int)g[u].size(); ++i) {int v = g[u][i];if (d[v] != -1) continue;d[v] = d[u] + 1;que.push(v);}} } long long solve(int x, int y) {for (int i = 1; i <= n; ++i) {dis[i] = D[0][x][i] + D[1][y][i];}for (int i = 1; i <= n+n; ++i) num[i] = 0;for (int i = 1; i <= n; ++i) num[dis[i]]++;for (int i = 1; i <= n+n; ++i) num[i] += num[i-1];for (int i = 1; i <= n; ++i) Rank[num[dis[i]]--] = i;queue<int> que1, que2;for (int i = 1; i <= n; ++i) que1.push(Rank[i]);while (!que1.empty() || !que2.empty()) {if (que2.empty() || (!que1.empty() && dis[que1.front()] < dis[que2.front()])) {int u = que1.front();que1.pop();for (int i = 0; i < (int)g[u].size(); ++i) {int v = g[u][i];if (dis[v] <= dis[u] + 1) continue;dis[v] = dis[u] + 1;que2.push(v);}}else {int u = que2.front();que2.pop();for (int i = 0; i < (int)g[u].size(); ++i) {int v = g[u][i];if (dis[v] <= dis[u] + 1) continue;dis[v] = dis[u] + 1;que2.push(v);}}}long long sum = 0;for (int i = 1; i <= n; ++i) sum += dis[i];return sum; } int main() {ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int T, Case = 1;scanf("%d", &T);while (T--) {printf("Case #%d: ", Case++);scanf("%d %d", &n, &m);for (int i = 1; i <= n; ++i) g[i].clear();for (int i = 0; i < m; ++i) {int u, v;scanf("%d %d", &u, &v);g[u].push_back(v);g[v].push_back(u);}scanf("%d", &a[0]);for (int i = 1; i <= a[0]; ++i) {scanf("%d", &a[i]);bfs(a[i], D[0][i]);}scanf("%d", &b[0]);for (int i = 1; i <= b[0]; ++i) {scanf("%d", &b[i]);bfs(b[i], D[1][i]);}long long P = 0;long long Q = a[0] * b[0] * n;for (int i = 1; i <= a[0]; ++i) {for (int j = 1; j <= b[0]; ++j) {P += solve(i, j);}}long long d = __gcd(P, Q);P /= d;Q /= d;if (Q == 1) printf("%lld", P);else printf("%lld/%lld", P, Q);puts("");}return 0; } // 做法2 #include <bits/stdc++.h> const int maxn = 1e5 + 5; const int inf = 0x3f3f3f3f; using namespace std; vector<int> g[maxn]; int D[2][30][maxn], a[30], b[30], dis[maxn]; int n, m; void bfs(int s, int *d, int x) { for (int i = 1; i <= x; ++i) {d[i] = -1;}d[s] = 0;queue<int> que;que.push(s);while (!que.empty()) {int u = que.front();que.pop();for (int i = 0; i < (int)g[u].size(); ++i) {int v = g[u][i];if (d[v] != -1) continue;d[v] = d[u] + 1;que.push(v);}} } long long solve(int x, int y) {int mx = 0;for (int i = 1; i <= n; ++i) mx = max(mx, D[0][x][i] + D[1][y][i]);for (int i = n+1; i <= mx+n+1; ++i) g[i].clear();for (int i = n+1; i <= mx+n; ++i) g[i].push_back(i+1);for (int i = 1; i <= n; ++i) g[n+1+D[0][x][i]+D[1][y][i]].push_back(i);bfs(n+1, dis, mx+n+1);long long sum = 0;for (int i = 1; i <= n; ++i) sum += dis[i];return sum; } int main() {ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int T, Case = 1;scanf("%d", &T);while (T--) {printf("Case #%d: ", Case++);scanf("%d %d", &n, &m);for (int i = 1; i <= n; ++i) g[i].clear();for (int i = 0; i < m; ++i) {int u, v;scanf("%d %d", &u, &v);g[u].push_back(v);g[v].push_back(u);}scanf("%d", &a[0]);for (int i = 1; i <= a[0]; ++i) {scanf("%d", &a[i]);bfs(a[i], D[0][i], n);}scanf("%d", &b[0]);for (int i = 1; i <= b[0]; ++i) {scanf("%d", &b[i]);bfs(b[i], D[1][i], n);}long long P = 0;long long Q = a[0] * b[0] * n;for (int i = 1; i <= a[0]; ++i) {for (int j = 1; j <= b[0]; ++j) {P += solve(i, j);}}long long d = __gcd(P, Q);P /= d;Q /= d;P -= Q;if (Q == 1) printf("%lld", P);else printf("%lld/%lld", P, Q);puts("");}return 0; } 與50位技術專家面對面20年技術見證，附贈技術全景圖

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