【POI2013】【BZOJ3417】Tales of seafaring

Description

Young Bytensson loves to hang out in the port tavern, where he often listens to the sea dogs telling their tales of seafaring. Initially, he believed them all, however incredible they sounded. Over time though, he became suspicious. He has decided to write a program that will verify if there may be any grain of truth in those tall stories. Bytensson reasoned that while he cannot tell if the sailors indeed weathered all those storms, he can at least find out if their travel itineraries make sense. This is a task for a programmer, which Bytensson, unfortunately, is not. Help him out!
There are ports and waterways connecting them in the waters frequented by the sailors Bytensson listened to. If there is a waterway between two ports, then sailing from one to the other is possible. Any waterway can be sailed in both directions.
Bytensson got to know K seafaring tales. Each tells of a sailor who began his journey in one port, sailed a number of waterways, and ended up in another port, which may have been the one he initially set sail from. The sailor in question may have sailed through the same waterway many times, each time in any direction.

一个n点m边无向图，边权均为1，有k个询问

每次询问给出(s,t,d)，要求回答是否存在一条从s到t的路径，长度为d

路径不必是简单路(可以自交)

Input

In the first line of the standard input, there are three integers, N,M and K (2<=N<=5000,1<=M<=5000,1<=K<=1000000) These denote, respectively: the number of ports in the waters frequented by the sailors who told Bytensson their stories, the number of waterways, and the number of tales.
The M lines that follow specify the waterways. A single waterway’s description consists of a single line that contains two integers, a and b (1<=a,b<=N,a<>b) separated by a single space; these specify the numbers of ports at the two ends of this particular waterway.
The K lines that follow specify the tales that Bytensson has heard. A single tale’s description consists of a single line with three integers, s,t and d (1<=S,T<=N,1<=d<=1000000000) separated by single spaces. These indicate that the tale’s protagonist set sail from port no. s, ended the journey in port no. t, and sailed exactly d times through various waterways.

Output

Your program should print exactly K lines to the standard output; the i-th of them should contain the word TAK (Polish for yes) if the journey described in the i-th tale (in input order) could have taken place. If it could not, then the line should contain the word NIE(Polish for no).

Sample Input

8 7 4

1 2

2 3

3 4

5 6

6 7

7 8

8 5

2 3 1

1 4 1

5 5 8

1 8 10

Sample Output

TAK

NIE

TAK

NIE

HINT

Source

鸣谢Wcmg提供译文

一开始理解错题目意思了,后来发现意思是这样:
对于一条边u,v,我们允许沿这条边从u走到v再走回u再走到v这样不停走
所以只要答案和最短路的奇偶性是一样的,就可以走出来

看了一下Claris代码发现精简而惊艳..可以把判断奇偶的数组和已有的部分数组合并起来,又短又快

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define MAXN 10010#define SIZE 1000010#define GET (ch>='0'&&ch<='9')using namespace std;int n,m,Q,top,tp;int dis[MAXN][2],pos[MAXN][2],q[MAXN][2],h,t;struct edge {   int to,w;   edge *next; }e[MAXN<<1],*prev[MAXN],E[SIZE],*Prev[SIZE];inline void insert1(int u,int v)    {   e[++top].to=v;e[top].next=prev[u];prev[u]=&e[top];  }inline void insert2(int u,int v,int w)  {   E[++tp].to=v;E[tp].next=Prev[u];Prev[u]=&E[tp];E[tp].w=w;   }inline void inque(int s,int f,int w,int now){    if (pos[s][f]<now)        pos[s][f]=now,dis[s][f]=w,        q[++t][0]=s,q[t][1]=f;}inline void in(int &x){    char ch=getchar();x=0;    while (!GET)    ch=getchar();    while (GET) x=x*10+ch-'0',ch=getchar();}int main(){    in(n);in(m);in(Q);int u,v,w;    while (m--) in(u),in(v),insert1(u,v),insert1(v,u);    while (Q--) in(u),in(v),in(w),insert2(u,v,w);    for (int x=1;x<=n;x++)    {        if (!Prev[x])   continue;        h=1;t=0;inque(x,0,0,x);        while (h<=t)        {            u=q[h][0];v=q[h++][1];            for (edge *i=prev[u];i;i=i->next)   inque(i->to,v^1,dis[u][v]+1,x);        }        for (edge *i=Prev[x];i;i=i->next)            if ((i->to!=x||prev[x])&&pos[i->to][i->w&1]==x&&dis[i->to][i->w&1]<=i->w)   i->to=0;    }    for (int i=1;i<=tp;i++) puts(E[i].to?"NIE":"TAK");}