逆向+两次bfs（UVA 1599）

  1 #include <iostream>
  2 #include <sstream>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <cmath>
  6 #include <string>
  7 #include <vector>
  8 #include <set>
  9 #include <cctype>
 10 #include <algorithm>
 11 #include <cmath>
 12 #include <deque>
 13 #include <queue>
 14 #include <map>
 15 #include <stack>
 16 #include <list>
 17 #include <iomanip>
 18 using namespace std;
 19 
 20 #define INF 0x7fffffff
 21 #define maxn 100000+10
 22 
 23 int n, m;
 24 vector<int> g[maxn];//用以存放互相连通的房间
 25 vector<int> col[maxn];//用以存放走廊颜色
 26 int step[maxn];//存放由n到i的最短步数
 27 int ans[maxn*2];
 28 int vis[maxn];
 29 void init()
 30 {
 31     for(int i = 0; i < maxn; i++)
 32     {
 33         g[i].clear();
 34         col[i].clear();
 35     }
 36     memset(step, -1, sizeof(step));
 37     memset(ans, 0, sizeof(ans));
 38     memset(vis, 0, sizeof(vis));
 39 }
 40 //==============获得到n的最少步数(逆向由n到1)===========
 41 void bfs1()
 42 {
 43     queue<int> q;
 44     q.push(n);
 45     step[n] = 0;//初始，由n到n的最短步数为0
 46     while(!q.empty())
 47     {
 48         int u = q.front(); q.pop();
 49         int sz = g[u].size();
 50         for(int i = 0; i < sz; i++)
 51         {
 52             int v = g[u][i];
 53 
 54             if(v == 1)
 55             {
 56                 step[v] = step[u]+1;
 57                 return ;
 58             }
 59 
 60             if(step[v] == -1)
 61             {
 62                 step[v] = step[u]+1;
 63                 q.push(v);
 64             }
 65         }
 66     }
 67     return ;
 68 }
 69 
 70 //==========获得最少步数时的最小走廊颜色===========
 71 void bfs2()
 72 {
 73     queue<int> q;
 74     q.push(1);
 75     while(!q.empty())
 76     {
 77         int u = q.front(); q.pop();
 78         ///
 79         if(!step[u])    return ;//到达n
 80         ///
 81         int mmin = INF;
 82         int sz = g[u].size();
 83         for(int i = 0; i < sz; i++)
 84         {
 85             int v = g[u][i];
 86             if(step[v] == step[u]-1)
 87             {
 88                 mmin = min(mmin, col[u][i]);//注意理解c[u][i]与g[u][i]间的联系--c[u][i]是u连接g[u][i]的走廊颜色
 89             }
 90         }
 91         //==========以上获得了从1出发最短路中每步的最小色
 92 
 93         int tmp_step = step[1] - step[u];//从1到u的步数，即出发第tmp_step步
 94         //ans[tmp_step] = (ans[tmp_step] == 0 ? mmin : min(mmin, ans[tmp_step]));
 95         if(ans[tmp_step] == 0)  ans[tmp_step] = mmin;
 96         else ans[tmp_step] = min(ans[tmp_step], mmin);
 97 
 98         for(int i = 0; i < sz; i++)
 99         {
100             int v = g[u][i];
101             ///该处判断条件很重要,把走过的路做标记
102             if(!vis[v] && step[v] == step[u]-1 && mmin == col[u][i])
103             {
104                 vis[v] = 1;
105                 q.push(v);
106             }
107         }
108     }
109     return ;
110 }
111 int main()
112 {
113     //===================input=====================
114     while(~scanf("%d%d", &n, &m))
115     {
116         init();
117         while(m--)
118         {
119             int a, b, c;
120             scanf("%d%d%d", &a, &b, &c);
121             if(a == b) continue;
122             g[a].push_back(b);
123             g[b].push_back(a);
124             col[a].push_back(c);
125             col[b].push_back(c);
126         }
127         //===============逆向bfs===============
128         //============获得最短步数=============
129         bfs1();
130         //===============正向bfs===============
131         //==========获得每步的走廊颜色=========
132         bfs2();
133 
134         printf("%d\n", step[1]);
135         for(int i = 0; i < step[1]; i++)
136         {
137             if(i) printf(" ");
138             printf("%d", ans[i]);
139         }
140         printf("\n");
141     }
142     return 0;
143 }