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大白书P330 这题比较麻烦
给出一个n个节点m条边的无向图,每条边上有一个正权。令c等于每对节点的最短路长度之和。例n=3时, c = d(1,1)+d(1,2)+d(1,3)+d(2,1)+d(2,2)+d(2,3)+d(3,1)+d(3,2)+d(3,3);
要求删除一条边后使得新的c值c‘最大。不连通的两点的最短路径长度为L
// LA4080/UVa1416 Warfare And Logistics // Rujia Liu #include<cstdio> #include<cstring> #include<vector> #include<algorithm> #include<queue> using namespace std; const int INF = 1000000000; 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]; // 是否已永久标号 int d[maxn]; // s到各个点的距离 int p[maxn]; // 最短路中的上一条弧 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> Q; for(int i = 0; i < n; i++) d[i] = INF; d[s] = 0; memset(done, 0, 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(e.dist > 0 && d[e.to] > d[u] + e.dist) { // 此处和模板不同,忽略了dist=-1的边。此为删除标记。根据题意和dijkstra算法的前提,正常的边dist>0 d[e.to] = d[u] + e.dist; p[e.to] = G[u][i]; Q.push((HeapNode){d[e.to], e.to}); } } } } }; //////// 题目相关 Dijkstra solver; int n, m, L; vector<int> gr[maxn][maxn]; // 两点之间的原始边权 int used[maxn][maxn][maxn]; // used[src][a][b]表示源点为src的最短路树是否包含边a->b int idx[maxn][maxn]; // idx[u][v]为边u->v在Dijkstra求解器中的编号 int sum_single[maxn]; // sum_single[src]表示源点为src的最短路树的所有d之和 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() { while(scanf("%d%d%d", &n, &m, &L) == 3) { 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 a, b, s; scanf("%d%d%d", &a, &b, &s); a--; b--; gr[a][b].push_back(s); gr[b][a].push_back(s); } // 构造网络 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(); int 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]; // 大二短边 c2 = max(c2, compute_newc(i, j)); e1 = e2 = gr[i][j][0]; // 恢复 } printf("%d %d\n", c, c2); } return 0; }
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原文地址:http://www.cnblogs.com/Opaser/p/4320533.html