Ultraviolet A 1416-warfare and logistics (Shortest Path Tree)

Va 1416-warfare and Logistics

Question Link

Given an undirected graph, each side has a positive weight, and C is equal to the sum of the shortest lengths of the two vertices. What is the maximum C value of the new graph after deleting one side?

Idea: It is too complicated to directly enumerate and delete edges. If the Shortest Path Tree is built, if the edges are deleted in the shortest path tree, they need to be done, in this way, the complexity is optimized to (n ^ 2 mlog (n), barely acceptable

Code:

#include <cstdio>#include <cstring>#include <vector>#include <queue>using namespace std;#define INF 0x3f3f3f3f3f3f3ftypedef long long ll;const int MAXNODE = 105;struct Edge {int u, v;ll dist;Edge() {}Edge(int u, int v, ll dist) {this->u = u;this->v = v;this->dist = dist;}};struct HeapNode {ll d;int u;HeapNode() {}HeapNode(ll d, int u) {this->d = d;this->u = u;}bool operator < (const HeapNode& c) const {return d > c.d;}};struct Dijkstra {int n, m;vector<Edge> edges;vector<int> g[MAXNODE];bool done[MAXNODE];ll d[MAXNODE];int p[MAXNODE];void init(int tot) {n = tot;for (int i = 0; i < n; i++)g[i].clear();edges.clear();}void add_Edge(int u, int v, ll dist) {edges.push_back(Edge(u, v, dist));m = edges.size();g[u].push_back(m - 1);}void print(int e) {//shun xuif (p[e] == -1) {printf("%d", e + 1);return;}print(edges[p[e]].u);printf(" %d", e + 1);}void print2(int e) {//ni xuif (p[e] == -1) {printf("%d\n", e + 1);return;}printf("%d ", e + 1);print2(edges[p[e]].u);}void dijkstra(int s) {priority_queue<HeapNode> Q;for (int i = 0; i < n; i++) d[i] = INF;d[s] = 0;p[s] = -1;memset(done, false, sizeof(done));Q.push(HeapNode(0, s));while (!Q.empty()) {HeapNode x = Q.top(); Q.pop();int u = x.u;if (done[u]) continue;done[u] = true;for (int i = 0; i < g[u].size(); i++) {Edge& e = edges[g[u][i]];if (d[e.v] > d[u] + e.dist) {d[e.v] = d[u] + e.dist;p[e.v] = g[u][i];Q.push(HeapNode(d[e.v], e.v));}}}}} gao, gao2;const int M = 1005;int n, m;ll L, c[M];Edge e[M];int main() {while (~scanf("%d%d%lld", &n, &m, &L)) {gao.init(n);int u, v;ll dist;for (int i = 0; i < m; i++) {scanf("%d%d%lld", &u, &v, &dist);u--; v--;e[i] = Edge(u, v, dist);gao.add_Edge(u, v, dist);gao.add_Edge(v, u, dist);}memset(c, 0, sizeof(c));ll ans1 = 0, ans2 = 0;for (int i = 0; i < n; i++) {gao.dijkstra(i);for (int j = 0; j < n; j++) {if (gao.d[j] == INF) ans1 += L;else ans1 += gao.d[j];}for (int j = 0; j < m; j++) {int u = e[j].u, v = e[j].v;if ((gao.p[v] != -1 && gao.edges[gao.p[v]].u == u) || ((gao.p[u] != -1) && gao.edges[gao.p[u]].u == v)) {gao2.init(n);for (int k = 0; k < m; k++) {if (j == k) continue;gao2.add_Edge(e[k].u, e[k].v, e[k].dist);gao2.add_Edge(e[k].v, e[k].u, e[k].dist);}gao2.dijkstra(i);for (int k = 0; k < n; k++) {if (gao2.d[k] == INF) c[j] += L;else c[j] += gao2.d[k];}} else {for (int k = 0; k < n; k++) {if (gao.d[k] == INF) c[j] += L;else c[j] += gao.d[k];}}}}for (int i = 0; i < m; i++)ans2 = max(ans2, c[i]);printf("%lld %lld\n", ans1, ans2);}return 0;}

Ultraviolet A 1416-warfare and logistics (Shortest Path Tree)