uva1416 dijkstra

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
", c, c2);
  }
  return 0;
}