Bellman-Ford || SPFA :Wormholes
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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Status
Description
While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.
As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .
To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.
Input
Line 1: A single integer,
F.
F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively:
N,
M, and
W
Lines 2..
M+1 of each farm: Three space-separated numbers (
S,
E,
T) that describe, respectively: a bidirectional path between
S and
E that requires
T seconds to traverse. Two fields might be connected by more than one path.
Lines
M+2..
M+
W+1 of each farm: Three space-separated numbers (
S,
E,
T) that describe, respectively: A one way path from
S to
E that also moves the traveler back
T seconds.
Output
Lines 1..
F: For each farm, output "YES"if FJ can achieve his goal, otherwise output "NO"(do not include the quotes).
Sample Input
2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8
Sample Output
NO
YES
Hint
For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.
이 문제의 첫인상은 Floyd로 해야지 죽도록 살아갈 수 없다
Bellman-Ford와 SPFA로 전환하여 데이터가 약합니다. 첫 번째 점을 원점으로 삼기만 하면 됩니다.
그리고 모든 초기화 문장을 꼭 기억해야 해요,wa는 수없이 찾아냈는데...
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
const int INF = 1<<31-1;
int n, m, w;
int d[550];
int cnt;
struct edge{
int u, v, w;
}e[6000];
bool bellman(){
int i, j, k;
for(i = 1; i <= n; i ++){
d[i] = INF;
}
d[1] = 0;
for(k = 0; k < n-1; k ++){
int flag = 1;//
for(i = 1; i <= cnt; i ++){
int x = e[i].u;
int y = e[i].v;
if(d[x] != INF)
if(d[y] > d[x] + e[i].w){
d[y] = d[x] + e[i].w;
flag = 0;
}
}
if(flag)
break;
}
for(i = 1; i <= cnt; i ++){
if(d[e[i].v] > d[e[i].u] + e[i].w)
return true; //
}
return false;
}
int main(){
int f;
int i;
cin >> f;
while(f -- ){
cnt = 0;
scanf("%d %d %d", &n, &m, &w);
for(i = 1; i <= m; i ++){
cnt ++;
scanf("%d %d %d", &e[cnt].u, &e[cnt].v, &e[cnt].w);
cnt ++;
e[cnt].u = e[cnt-1].v;
e[cnt].v = e[cnt-1].u;
e[cnt].w = e[cnt-1].w;
}
for(i = 1; i <= w; i ++){
cnt ++;
scanf("%d %d %d", &e[cnt].u, &e[cnt].v, &e[cnt].w);
e[cnt].w = -e[cnt].w;
}
if(bellman())
printf("YES
");
else
printf("NO
");
}
return 0;
}
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
using namespace std;
const int INF = 1<<31-1;
int n, m, ww;
int d[55000];
int cnt;
struct node{
int to;
int next;// , next
int weight;
}e[60000];
int lastshow[40000], num[40000];
bool inqueue[40000];
queue<int>q;
void insert(int a, int b, int w){
e[++cnt].to = b;
e[cnt].weight = w;
e[cnt].next = lastshow[a];
lastshow[a] = cnt;//lastshow , next lastshow
}
bool spfa(){
q.push(1);
num[1] ++;
while(!q.empty()){
int x = q.front();
q.pop();
inqueue[x] = false;
int id = lastshow[x];
while(id != -1){
if(d[x] < INF && d[e[id].to] > e[id].weight + d[x]){
d[e[id].to] = e[id].weight + d[x];
if(!inqueue[e[id].to]){
inqueue[e[id].to] = true;
q.push(e[id].to);
}
num[id]++;
if( num[id] > n )// ,
return true;
}
id = e[id].next;
}
}
return false;
}
int main(){
int f;
int i;
int a, b, w;
cin >> f;
while(f -- ){
cnt = 0;
scanf("%d %d %d", &n, &m, &ww);
memset(lastshow,-1,sizeof(lastshow));
memset(inqueue, false, sizeof(inqueue));
memset(num, 0, sizeof(num));
for(i = 1; i <= n; ++ i)
d[i]=INF;
d[1]=0;
cnt=0;
while(!q.empty())
{
q.pop();
}
for(i = 1; i <= m; i ++){
scanf("%d %d %d", &a, &b, &w);
insert(a, b, w);
insert(b, a, w);
}
for(i = 1; i <= ww; i ++){
scanf("%d %d %d", &a, &b, &w);
insert(a, b, -w);
}
if(spfa())
printf("YES
");
else
printf("NO
");
}
return 0;
}
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