[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：

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

转载于:https://www.cnblogs.com/whistle13326/p/7406338.html