[USACO09JAN]安全出行Safe Travel
题目描述
Gremlins have infested the farm. These nasty, ugly fairy-like
creatures thwart the cows as each one walks from the barn (conveniently located at pasture_1) to the other fields, with cow_i traveling to from pasture_1 to pasture_i. Each gremlin is personalized and knows the quickest path that cow_i normally takes to pasture_i. Gremlin_i waits for cow_i in the middle of the final cowpath of the quickest route to pasture_i, hoping to harass cow_i.
Each of the cows, of course, wishes not to be harassed and thus chooses an at least slightly different route from pasture_1 (the barn) to pasture_i.
Compute the best time to traverse each of these new not-quite-quickest routes that enable each cow_i that avoid gremlin_i who is located on the final cowpath of the quickest route from pasture_1 to
pasture_i.
As usual, the M (2 <= M <= 200,000) cowpaths conveniently numbered 1..M are bidirectional and enable travel to all N (3 <= N <= 100,000) pastures conveniently numbered 1..N. Cowpath i connects pastures a_i (1 <= a_i <= N) and b_i (1 <= b_i <= N) and requires t_i (1 <= t_i <= 1,000) time to traverse. No two cowpaths connect the same two pastures, and no path connects a pasture to itself (a_i != b_i). Best of all, the shortest path regularly taken by cow_i from pasture_1 to pasture_i is unique in all the test data supplied to your program.
By way of example, consider these pastures, cowpaths, and [times]:
1--[2]--2-------+
| | | [2] [1] [3]
| | | +-------3--[4]--4
TRAVEL BEST ROUTE BEST TIME LAST PATH
p_1 to p_2 1->2 2 1->2
p_1 to p_3 1->3 2 1->3
p_1 to p_4 1->2->4 5 2->4
When gremlins are present:
TRAVEL BEST ROUTE BEST TIME AVOID
p_1 to p_2 1->3->2 3 1->2
p_1 to p_3 1->2->3 3 1->3
p_1 to p_4 1->3->4 6 2->4
For 20% of the test data, N <= 200.
For 50% of the test data, N <= 3000.
TIME LIMIT: 3 Seconds
MEMORY LIMIT: 64 MB
Gremlins最近在农场上泛滥,它们经常会阻止牛们从农庄(牛棚_1)走到别的牛棚(牛_i的目的 地是牛棚_i).每一个gremlin只认识牛_i并且知道牛_i一般走到牛棚_i的最短路经.所以它 们在牛_i到牛棚_i之前的最后一条牛路上等牛_i. 当然,牛不愿意遇到Gremlins,所以准备找 一条稍微不同的路经从牛棚_1走到牛棚_i.所以,请你为每一头牛_i找出避免gremlin_i的最 短路经的长度. 和以往一样, 农场上的M (2 <= M <= 200,000)条双向牛路编号为1..M并且能让所有牛到 达它们的目的地, N(3 <= N <= 100,000)个编号为1..N的牛棚.牛路i连接牛棚a_i (1 <= a_i <= N)和b_i (1 <= b_i <= N)并且需要时间t_i (1 <=t_i <= 1,000)通过. 没有两条牛路连接同样的牛棚,所有牛路满足a_i!=b_i.在所有数据中,牛_i使用的牛棚_1到牛 棚_i的最短路经是唯一的.
输入输出格式
输入格式：
- Line 1: Two space-separated integers: N and M
- Lines 2..M+1: Three space-separated integers: a_i, b_i, and t_i
输出格式：
- Lines 1..N-1: Line i contains the smallest time required to travel from pasture_1 to pasture_i+1 while avoiding the final cowpath of the shortest path from pasture_1 to pasture_i+1. If no such path exists from pasture_1 to pasture_i+1, output -1 alone on the line.
输入输出样例
输入样例#1：
4 5 1 2 2 1 3 2 3 4 4 3 2 1 2 4 3
输出样例#1：
3 3 6 先构出最短路径树（题目说唯一）对于一个非树边(u,v) 一定使树构成一个环并且使环上的点（除了lca(u,v)）都有了一个次短路对于点i 这个所谓次短路的长度就是dist[u]+dist[v]+value[u,v]-dist[i]可以发现次短路大小长度跟u,v有关 跟i无关所以对于一个非树边（u,v)设它的权值为dist[u]+dist[v]+value[u,v]把这个权值按从小到大排序 每次把环上没被赋值过的点赋上值快速实现上面那个 可以用并查集
1 #include <queue>
2 #include <cstdio>
3 #include <ctype.h>
4 #include <algorithm>
5
6 using namespace std;
7
8 const int MAXN=100010;
9
10 int n,m;
11
12 struct node {
13 int to;
14 int next;
15 int val;
16 };
17 node e[MAXN*10];
18
19 struct othertree{
20 int u,v;
21 int val;
22 bool operator < (const othertree &h) const {
23 return val<h.val;
24 }
25 };
26 othertree tr[MAXN*5];
27
28 struct fuck {
29 int to,cost;
30 fuck(int a,int b): to(a),cost(b) {}
31 bool operator < (const fuck&k) const {
32 return cost>k.cost;
33 }
34 };
35
36 int head[MAXN],tot,inr;
37
38 int dis[MAXN],dep[MAXN],fa[MAXN],Id[MAXN],ans[MAXN],da[MAXN];
39
40 bool vis[MAXN],inthetree[MAXN<<1];
41
42 priority_queue<fuck> q;
43
44 inline void add(int x,int y,int z) {
45 e[++tot].to=y;
46 e[tot].val=z;
47 e[tot].next=head[x];
48 head[x]=tot;
49 }
50
51 inline void read(int&x) {
52 int f=1;register char c=getchar();
53 for(x=0;!isdigit(c);c==‘-‘&&(f=-1),c=getchar());
54 for(;isdigit(c);x=x*10+c-48,c=getchar());
55 x=x*f;
56 }
57
58 inline void dij() {
59 q.push(fuck(1,0));
60 for(int i=1;i<=n;++i) dis[i]=1e9+1e9;
61 dis[1]=0;dep[1]=1;
62 while(!q.empty()) {
63 while(!q.empty()&&vis[q.top().to]) q.pop();
64 fuck p=q.top();
65 if(vis[p.to]) continue;
66 vis[p.to]=true;
67 for(int i=head[p.to];i;i=e[i].next) {
68 int to=e[i].to;
69 if(dis[to]>dis[p.to]+e[i].val&&!vis[to]) {
70 dis[to]=dis[p.to]+e[i].val;
71 fa[to]=p.to;
72 dep[to]=dep[p.to]+1;
73 q.push(fuck(to,dis[to]));
74 }
75 }
76 }
77 return;
78 }
79
80 inline void swap(int&a,int&b) {
81 int t=a;a=b;b=t;
82 return;
83 }
84
85 inline int find(int x) {
86 return da[x]==x?x:da[x]=find(da[x]);
87 }
88
89 inline void Ask_ans(int x,int y,int v) {
90 int lastx=-1,lasty=-1;
91 while(find(x)!=find(y)) {
92 if(dep[find(x)]<dep[find(y)]) swap(x,y),swap(lastx,lasty);
93 if(ans[x]==-1) {
94 ans[x]=v-dis[x];
95 if(lastx!=-1) da[lastx]=x;
96 }
97 else if(lastx!=-1)
98 da[lastx]=find(x);
99 lastx=find(x);
100 x=fa[lastx];
101 }
102 return;
103 }
104
105 int hh() {
106 freopen("travel.in","r",stdin);
107 freopen("travel.out","w",stdout);
108 int x,y,z;
109 read(n);read(m);
110 for(int i=1;i<=m;++i)
111 read(x),read(y),read(z),add(x,y,z),add(y,x,z);
112 dij();
113 for(int i=1;i<=n;++i)
114 for(int j=head[i];j;j=e[j].next){
115 int to=e[j].to;
116 if(fa[to]==i||fa[i]==to) continue;
117 tr[inr].u=i;tr[inr].v=to;
118 tr[inr].val=e[j].val+dis[i]+dis[to];
119 ++inr;
120 }
121 sort(tr,tr+inr);
122 for(int i=1;i<=n;++i) da[i]=i,ans[i]=-1;
123 for(int i=0;i<inr;++i)
124 Ask_ans(tr[i].u,tr[i].v,tr[i].val);
125 for(int i=2;i<=n;++i) printf("%d\n",ans[i]);
126 return 0;
127 }
128
129 int sb=hh();
130 int main() {;}
代码