LA4080 Warfare And Logistics

微微发亮的传送门

又来一弹白书例题，发现……往往是不看题解我是做不出来题的。。sad……

首先这道题很容易让人一冲动就用floyd来解了，可是，明显folyd的复杂度我们是无法接受的，100个结点加上1000条边，一定会超时的，那么有关最短路的就剩下一个spfa和dijkstra了，其实dijkstra或者spfa对于每一个起点，都能求出来一棵树，暂且称它为最短路树，对于这课树上的结点，如果我们不破坏这棵树上的边，那么就不用重新求任何一条最短路，如果破坏了其中一条边，那只需要对于那条边对应的根结点重新求一次就行了，注意重边的处理就行了。

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>using namespace std;const int INF = 0x7fffffff;const int maxn = 100 + 10;struct Edge{int from, to, dist;};struct HeapNode{int d, u;bool operator < (const HeapNode& rhs) const {return d > rhs.d;}};struct Dijkstra{int n, m;vector<Edge> edges;vector<int> G[maxn];bool done[maxn];int p[maxn], d[maxn];void init(int n){this -> n = n;for (int i = 0; i < n; i++)G[i].clear();edges.clear();}void AddEdge(int from, int to, int dist){edges.push_back((Edge){from, to, dist});m = edges.size();G[from].push_back(m - 1);}void dijkstra(int s){priority_queue<HeapNode> que;for (int i = 0; i < n; i++)d[i] = INF;d[s] = 0;memset(done, 0, sizeof(done));que.push((HeapNode){0, s});while(!que.empty()){HeapNode x = que.top(); que.pop();int u = x.u;if (done[u]) continue;done[u] = 1;for (int i = 0; i < G[u].size(); i++){Edge& e = edges[G[u][i]];if (e.dist >= 0 && d[e.to] > d[u] + e.dist){d[e.to] = d[u] + e.dist;p[e.to] = G[u][i];que.push((HeapNode){d[e.to], e.to});}}}}};Dijkstra solver;int n, m, L;vector<int> gr[maxn][maxn];int used[maxn][maxn][maxn];int idx[maxn][maxn];int sum_single[maxn];int compute_c(){int ans = 0;memset(used, 0, sizeof(used));for (int src = 0; src < n; src++){solver.dijkstra(src);sum_single[src] = 0;for (int i = 0; i < n; i++){if (i != src){int fa = solver.edges[solver.p[i]].from;used[src][fa][i] = used[src][i][fa] = 1;}sum_single[src] += (solver.d[i] == INF ? L : solver.d[i]);}ans += sum_single[src];}return ans;}int compute_newc(int a, int b){int ans = 0;for (int src = 0; src < n; src++)if (!used[src][a][b]) ans += sum_single[src];else{solver.dijkstra(src);for (int i = 0; i < n; i++)ans += (solver.d[i] == INF ? L : solver.d[i]);}return ans;}int main(){//freopen("in.txt", "r", stdin);while(~scanf("%d%d%d", &n, &m, &L)){solver.init(n);for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)gr[i][j].clear();for (int i = 0; i < m; i++){int x, y, z;scanf("%d%d%d", &x, &y, &z);x -= 1; y -= 1;gr[x][y].push_back(z);gr[y][x].push_back(z);}for (int i = 0; i < n; i++)for (int j = i + 1; j < n; j++)if (!gr[i][j].empty()){sort(gr[i][j].begin(), gr[i][j].end());solver.AddEdge(i, j, gr[i][j][0]);idx[i][j] = solver.m - 1;solver.AddEdge(j, i, gr[i][j][0]);idx[j][i] = solver.m - 1;}int c = compute_c(), c2 = -1;for (int i = 0; i < n; i++)for (int j = i + 1; j < n; j++)if (!gr[i][j].empty()){int& e1 = solver.edges[idx[i][j]].dist;int& e2 = solver.edges[idx[j][i]].dist;if (gr[i][j].size() == 1) e1 = e2 = -1;else e1 = e2 = gr[i][j][1];c2 = max(c2, compute_newc(i, j));e1 = e2 = gr[i][j][0];}printf("%d %d\n", c, c2);}return 0;}