BZOJ 2125: 最短路

2125: 最短路

Time Limit: 1 Sec  Memory Limit: 259 MB
Submit: 756  Solved: 331
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Description

给一个N个点M条边的连通无向图,满足每条边最多属于一个环,有Q组询问,每次询问两点之间的最短路径。

Input

输入的第一行包含三个整数,分别表示N和M和Q 下接M行,每行三个整数v,u,w表示一条无向边v-u,长度为w 最后Q行,每行两个整数v,u表示一组询问

Output

输出Q行,每行一个整数表示询问的答案

Sample Input

9 10 2
1 2 1
1 4 1
3 4 1
2 3 1
3 7 1
7 8 2
7 9 2
1 5 3
1 6 4
5 6 1
1 9
5 7

Sample Output

5
6

HINT

 

对于100%的数据,N<=10000,Q<=10000

 

Source

 
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  1 #include <cstdio>
  2 #include <cstring>
  3 
  4 inline int nextChar(void)
  5 {
  6     static const int siz = 1 << 20;
  7     
  8     static char buf[siz];
  9     static char *hd = buf + siz;
 10     static char *tl = buf + siz;
 11     
 12     if (hd == tl)
 13         fread(hd = buf, 1, siz, stdin);
 14         
 15     return int(*hd++);
 16 }
 17 
 18 inline int nextInt(void)
 19 {
 20     int r = 0, c = nextChar();
 21     
 22     for (; c < 48; c = nextChar());
 23     for (; c > 47; c = nextChar())
 24         r = r * 10 + c - '0';
 25     
 26     return r;
 27 }
 28 
 29 template <class T>
 30 inline T min(const T &a, const T &b)
 31 {
 32     return a < b ? a : b;
 33 }
 34 
 35 template <class T>
 36 inline T abs(const T &x)
 37 {
 38     return x < 0 ? -x : x;
 39 }
 40 
 41 const int mxn = 500005;
 42 const int inf = 0x3f3f3f3f;
 43 
 44 int n, m, q;
 45 
 46 int tot;
 47 int hd[mxn];
 48 int to[mxn];
 49 int vl[mxn];
 50 int nt[mxn];
 51 
 52 inline void add(int x, int y, int w)
 53 {
 54     nt[tot] = hd[x];
 55     to[tot] = y;
 56     vl[tot] = w;
 57     hd[x] = tot++;
 58 }
 59 
 60 int dis[mxn];
 61 int que[mxn];
 62 int vis[mxn];
 63 
 64 inline void spfa(void)
 65 {
 66     memset(vis,   0, sizeof vis);
 67     memset(dis, inf, sizeof dis);
 68     
 69     int lt = 0, rt = 0;
 70     
 71     que[rt++] = 1;
 72     vis[1] = 1;
 73     dis[1] = 0;
 74     
 75     while (lt != rt)
 76     {
 77         int u = que[lt++], v;
 78 
 79         if (lt > mxn)lt = 0;
 80         
 81         vis[u] = 0;
 82         
 83         for (int i = hd[u]; ~i; i = nt[i])
 84             if (v = to[i], dis[v] > dis[u] + vl[i])
 85             {
 86                 dis[v] = dis[u] + vl[i];
 87                 
 88                 if (!vis[v])
 89                 {
 90                     vis[que[rt++] = v] = 1;
 91 
 92                     if (rt > mxn)rt = 0;
 93                 }
 94             }
 95     }
 96 }
 97 
 98 struct data
 99 {
100     int u, v, w;
101     
102     inline data(void) {};
103     inline data(int a, int b, int c)
104         : u(a), v(b), w(c) {};
105 }stk[mxn]; int top;
106 
107 int tim;
108 int dfn[mxn];
109 int low[mxn];
110 
111 int cnt;
112 int cir[mxn];
113 int len[mxn];
114 int bel[mxn];
115 int fat[mxn][25];
116 
117 inline void addcir(int u, int v)
118 {
119     ++cnt;
120     
121     while (u != stk[top].u && v != stk[top].v)
122     {
123         int x = stk[top].u, y = stk[top].v, w = stk[top].w;
124         
125         if (x != u)bel[x] = cnt, fat[x][0] = u;
126         if (y != u)bel[y] = cnt, fat[y][0] = u;
127         
128         len[cnt] += w, cir[x] = cir[y] + w;
129         
130         --top;
131     }
132     
133     {
134         int x = stk[top].u, y = stk[top].v, w = stk[top].w;
135         
136         len[cnt] += w, cir[x] = cir[y] + w;
137         
138         fat[y][0] = x;
139         
140         --top;
141     }
142 }
143 
144 void tarjan(int u, int f)
145 {
146     dfn[u] = low[u] = ++tim;
147     
148     for (int i = hd[u], v; ~i; i = nt[i])
149         if ((v = to[i]) != f)
150         {
151             if (!dfn[v])
152             {
153                 stk[++top] = data(u, v, vl[i]), tarjan(v, u);
154                 
155                 if (low[v] < low[u])low[u] = low[v];
156                 
157                 if (low[v] >= dfn[u])addcir(u, v);
158             }
159             else if (dfn[v] < low[u])
160                 low[u] = dfn[v], stk[++top] = data(u, v, vl[i]);    
161         }
162 }
163 
164 int dep[mxn];
165 
166 void build(int u)
167 {
168     for (int i = hd[u]; ~i; i = nt[i])
169         dep[to[i]] = dep[u] + 1, build(to[i]);
170 }
171 
172 inline void prework(void)
173 {
174     spfa();
175     
176     tarjan(1, 0);
177     
178     for (int i = 1; i <= 20; ++i)
179         for (int j = 1; j <= n; ++j)
180             fat[j][i] = fat[fat[j][i - 1]][i - 1];
181     
182     memset(hd, -1, sizeof hd), tot = 0;
183     
184     for (int i = 2; i <= n; ++i)
185         add(fat[i][0], i, 0);
186     
187     dep[1] = 1, build(1);
188 }
189 
190 inline int getLCA(int a, int b, int &x, int &y)
191 {
192     if (dep[a] < dep[b])
193         a ^= b ^= a ^= b;
194     
195     int ret = dis[a] + dis[b];
196     
197     for (int i = 20; ~i; --i)
198         if (dep[fat[a][i]] >= dep[b])
199             a = fat[a][i];
200     
201     if (a == b)
202         return x = y = a, ret - dis[a] * 2;
203     
204     for (int i = 20; ~i; --i)
205         if (fat[a][i] != fat[b][i])
206             a = fat[a][i],
207             b = fat[b][i];
208     
209     return x = a, y = b, ret - dis[fat[a][0]] * 2;
210 }
211 
212 signed main(void)
213 {
214     n = nextInt();
215     m = nextInt();
216     q = nextInt();
217     
218     memset(hd, -1, sizeof(hd)), tot = 0;
219     
220     for (int i = 1; i <= m; ++i)
221     {
222         int x = nextInt();
223         int y = nextInt();
224         int w = nextInt();
225         
226         add(x, y, w);
227         add(y, x, w);
228     }
229     
230     prework();
231     
232     for (int i = 1, x, y; i <= q; ++i)
233     {
234         int a = nextInt();
235         int b = nextInt();
236         
237         int ans = getLCA(a, b, x, y);
238         
239         if (bel[x] && bel[x] == bel[y])
240         {
241             ans = dis[a] + dis[b] - dis[x] - dis[y];
242             
243             int len1 = abs(cir[x] - cir[y]);
244             int len2 = len[bel[x]] - len1;
245             
246             ans += min(len1, len2);
247         }
248         
249         printf("%d\n", ans);
250     }
251 }

 

@Author: YouSiki

 

posted @ 2017-02-25 16:19  YouSiki  阅读(...)  评论(...编辑  收藏