为什么都说简单好想咧。坦白从宽看了人家的代码，涨了好多姿势，，
http://blog.csdn.net/u013382399/article/details/38227917
被一个细节坑了。。
2147483647是0x7fffffff啊啊啊，7个f！！！
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 ", step[1]); 135 for(int i = 0; i < step[1]; i++) 136 { 137 if(i) printf(" "); 138 printf("%d", ans[i]); 139 } 140 printf(" "); 141 } 142 return 0; 143 }