UVA1416/LA4080 Warfare And Logistics

题目大意：有N个点，M条路，如果两条路不连通的话，就将这两条路的距离设置为L
现在要求你求出每两点之间的最短距离和
接着要求
求出炸断 给出的M条路中的一条路后，每两点之间的最短距离和的最大值（翻译来自http://blog.csdn.net/l123012013048/article/details/47297393）

单源最短路树：把源点到其他点的最短路拼起来，形成最短路树（可能有多棵，这里只需要一棵）。我们把起点为i的单源最短路树称为i源最短路树。想要让最短路改变，删除的边必定在这课最短路树上（但反过来不成立，即删除最短路树上的边最短路改变，是错误的）。处理出单源最短路树，然后用belong[i][j]表示边i是否在j源最短路树上，枚举边，重新做belong[i][j]为1的点即可。

这里仅仅是带来常数上的优化，实际复杂度并不能改变（nm^2logn），蓝书说复杂度变了(n^2mlogn)。。我不敢苟同

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <algorithm>
  6 #include <queue>
  7 #include <vector>
  8 #include <map>
  9 #include <string>
 10 #include <cmath>
 11 #define min(a, b) ((a) < (b) ? (a) : (b))
 12 #define max(a, b) ((a) > (b) ? (a) : (b))
 13 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
 14 template<class T>
 15 inline void swap(T &a, T &b)
 16 {
 17     T tmp = a;a = b;b = tmp;
 18 }
 19 inline void read(int &x)
 20 {
 21     x = 0;char ch = getchar(), c = ch;
 22     while(ch < ‘0‘ || ch > ‘9‘) c = ch, ch = getchar();
 23     while(ch <= ‘9‘ && ch >= ‘0‘) x = x * 10 + ch - ‘0‘, ch = getchar();
 24     if(c == ‘-‘) x = -x;
 25 }
 26 const int INF = 0x3f3f3f3f;
 27 const int MAXN = 200 + 10;
 28 const int MAXM = 2000 + 10;
 29 struct Edge
 30 {
 31     int u,v,w,nxt;
 32     Edge(int _u, int _v, int _w, int _nxt){u = _u, v = _v, w = _w, nxt = _nxt;}
 33     Edge(){}
 34 }edge[MAXM << 1];
 35 int head[MAXN], cnt = 1, d[MAXN][MAXN], dis[MAXM], flag, belong[MAXM][MAXN], vis[MAXN], p[MAXN][MAXN], l, tmp1, tmp2, tmp3, n, m;
 36 //p[i]表示以i为起点的最短路树，在最短路树上，j的父亲的那条边
 37 inline void insert(int a, int b, int c)
 38 {
 39     edge[++ cnt] = Edge(a,b,c,head[a]), head[a] = cnt;
 40 }
 41 struct Node
 42 {
 43     int u,w;
 44     Node(int _u, int _w){u = _u, w = _w;}
 45     Node(){}
 46 };
 47 struct cmp
 48 {
 49     bool operator()(Node a, Node b)
 50     {
 51         return a.w > b.w;
 52     }
 53 };
 54 std::priority_queue<Node, std::vector<Node>, cmp> q;
 55 void dij(int S, int *d, int size)
 56 {
 57     while(q.size())q.pop();memset(d, 0x3f, size), d[S] = 0, memset(vis, 0, sizeof(vis)), q.push(Node(S, 0));
 58     while(q.size())
 59     {
 60         Node now = q.top();q.pop();
 61         if(vis[now.u]) continue; vis[now.u] = 1;
 62         for(int pos = head[now.u];pos;pos = edge[pos].nxt)
 63         {
 64             int v = edge[pos].v;
 65             if(d[v] > d[now.u] + edge[pos].w)
 66             {
 67                 d[v] = d[now.u] + edge[pos].w, q.push(Node(v, d[v]));
 68                 if(flag) p[S][v] = pos;
 69             }
 70         }
 71     }
 72 }
 73 int main()
 74 {
 75     while(scanf("%d %d %d", &n, &m, &l) != EOF)
 76     {
 77         flag = 1, memset(head, 0, sizeof(head)), memset(belong, 0, sizeof(belong)), memset(p, 0, sizeof(p)), cnt = 1;
 78         for(int i = 1;i <= m;++ i) read(tmp1), read(tmp2), read(tmp3), insert(tmp1, tmp2, tmp3), insert(tmp2, tmp1, tmp3);
 79         for(int i = 1;i <= n;++ i) dij(i, d[i], sizeof(d[i]));
 80         long long ans1 = 0, ans2 = ans1, ans3 = 0;
 81         for(int i = 1;i <= n;++ i)
 82             for(int j = 1;j <= n;++ j)
 83                 if(d[i][j] >= INF) ans1 += l;
 84                 else ans1 += d[i][j];
 85         printf("%lld ", ans1);
 86         flag = ans3 = 0;
 87         for(int i = 1;i <= n;++ i)
 88             for(int j = 1;j <= n;++ j)
 89                 if(p[i][j]) belong[p[i][j]][i] = 1;
 90         for(int pos = 2;pos <= cnt;pos += 2)
 91         {
 92             tmp1 = edge[pos].w;ans2 = ans1;
 93             edge[pos].w = edge[pos ^ 1].w = INF;
 94             for(int S = 1;S <= n;++ S)
 95                 if(belong[pos][S] || belong[pos^1][S])
 96                 {
 97                     dij(S, dis, sizeof(dis));
 98                     for(int i = 1;i <= n;++ i)
 99                     {
100                         if(d[S][i] >= INF) ans2 -= l;
101                         else ans2 -= d[S][i];
102                         if(dis[i] >= INF) ans2 += l;
103                         else ans2 += dis[i];
104                     }
105                 }
106             edge[pos].w = edge[pos ^ 1].w = tmp1;
107             ans3 = max(ans3, ans2);
108         }
109         printf("%lld\n", ans3);
110     }
111     return 0;
112 }

UVA1416/LA4080

原文地址：https://www.cnblogs.com/huibixiaoxing/p/8397479.html