Description
7 7 3 1 2 1 7 6 6 3 5 3 4 3 7 2 3 7 1 1 2 3 5 3 6 2 4 2 4 5 3 10 6 1 20 3 5 6 7
10 22 5
Solution
法一
法二
法三
法四
rethink
思路
考场上只想到暴力。。。
对于每一种可能的边，全部暴力建边。
注意建的是反向边
从1号节点（首都）再跑一遍最短路即可
时间复杂度
建边的时候对于好看的二叉树是n log n
链的话可能到 n^2
居然没卡我
再就是 dij 跟 spfa 了
最开始跑了spfa（打得快嘛）
最后要结束的时候发现自己建的边不就是稠密图嘛
马上换成 dij
但不知道为什么 spfa 还是比 dij 快了不少
（spfa）
（dij）
空间复杂度
怕炸空间（重建的边很多）只开了500W
结果最后一个点RE了。
mlg开了510W过了。。。
事实证明开了600W还稳得一批
code
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<queue> 7 #include<vector> 8 #include<stack> 9 #include<set> 10 #include<deque> 11 #include<map> 12 using namespace std; 13 14 template <typename T> void read(T &x) { 15 x = 0; int f = 1; char c; 16 for (c = getchar(); c < '0' || c > '9'; c = getchar()) if (c == '-') f = -f; 17 for (; c >= '0' && c <= '9'; c = getchar()) x = 10 * x + c - '0' ; 18 x *= f; 19 } 20 template <typename T> void write(T x){ 21 if (x < 0) putchar('-'), x = -x; 22 if (x > 9) write(x / 10); 23 putchar(x % 10 + '0'); 24 } 25 template <typename T> void writeln(T x) { write(x); putchar(' '); } 26 template <typename T> void writesn(T x) { write(x); putchar(' '); } 27 28 #define ll long long 29 #define inf 100000 30 #define next net 31 #define P 9999991 32 #define N 100010 33 #define mid ((l+r)>>1) 34 #define lson (o<<1) 35 #define rson (o<<1|1) 36 #define R register 37 #define debug puts("zxt") 38 39 int n, m , f[N ], q; 40 int cut, head[N ], next[N ], ver[N ]; 41 int cuta, heada[N ], nexta[N * 60], vera[N * 60], wa[N * 60]; 42 int dis[N ], book[N ]; 43 struct node{ 44 vector<int>k, w ; 45 }fee[N ]; 46 inline void add(int x, int y) 47 { 48 ver[++cut] = y; next[cut] = head[x]; head[x] = cut; 49 } 50 inline void add_again(int x, int y, int z) 51 { 52 vera[++cuta] = y; nexta[cuta] = heada[x]; heada[x] = cuta; wa[cuta] = z; 53 } 54 inline void dfs(int x, int fa) 55 { 56 f[x] = fa; 57 for(R int i = head[x]; i; i = next[i]) 58 { 59 int y = ver[i]; 60 if(y == fa) continue; 61 dfs(y, x); 62 } 63 for(R int i = 0; i < fee[x].k.size(); i++) 64 { 65 int kkk = x, num = fee[x].k[i]; 66 while(num && f[kkk]) 67 { 68 kkk = f[kkk]; num --; 69 add_again(kkk, x, fee[x].w [i]); 70 //writesn(kkk);writesn(x);writeln(fee[x].w[i]); 71 } 72 } 73 } 74 inline void spfa() 75 { 76 queue<int>q; 77 memset(dis,0x3f,sizeof(dis)); 78 q.push(1); 79 dis[1] = 0; 80 while(!q.empty()) 81 { 82 int x = q.front(); 83 q.pop(); 84 book[x] = 0; 85 for(R int i = heada[x]; i; i = nexta[i]) 86 { 87 int y = vera[i], z = wa[i]; 88 if(dis[y] > dis[x] + z) 89 { 90 dis[y] = dis[x] + z; 91 if(!book[y]) 92 { 93 book[y] = 1; 94 q.push(y); 95 } 96 } 97 } 98 } 99 } 100 inline void dij() 101 { 102 priority_queue<pair<int,int> >q; 103 memset(dis,0x3f,sizeof(dis)); 104 q.push(make_pair(0, 1)); 105 dis[1] = 0; 106 while(!q.empty()) 107 { 108 int x = q.top().second; 109 q.pop(); 110 if(book[x]) continue; 111 book[x] = 1; 112 for(R int i = heada[x]; i; i = nexta[i]) 113 { 114 int y = vera[i], z = wa[i]; 115 if(dis[y] > dis[x] + z) 116 { 117 dis[y] = dis[x] + z; 118 q.push(make_pair(-dis[y], y)); 119 } 120 } 121 } 122 } 123 signed main() 124 { 125 read(n); read(m ); 126 for(R int i = 1, x, y; i < n; i++) 127 { 128 read(x); read(y); 129 add(y, x); 130 } 131 for(R int i = 1, x, y; i <= m ; i++) 132 { 133 read(x); 134 read(y); 135 fee[x].k.push_back(y); 136 read(y); 137 fee[x].w.push_back(y); 138 } 139 dfs(1, 0); 140 dij(); //spfa(); 141 read(q); 142 while(q--) 143 { 144 int ask; 145 read(ask); 146 writeln(dis[ask]); 147 } 148 //for(R int i = 1; i <= n; i++)writesn(dis[i]); 149 return 0; 150 }