[题解] poj 3259 Wormholes (dfs/bellmanford判负环)
- 传送门 -
http://poj.org/problem?id=3259
| Time Limit: 2000MS | | Memory Limit: 65536K |
| Total Submissions: 54401 | | Accepted: 20260 |
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.
Source
- 题意 -
给定n个节点, m条无向边, w条有向边.
无向边权值为正, 有向边权值为负.(输入中都为正, 有向边权值要取反)
判断有没有负环.
- 思路 -
把无向边改为两条有向边即可.
之前在类似题目中用到了dfs和bfs, 但是上次用bfs很慢, 这次就没有用了.
试了一下bellmanford, 跑得很快, 尤其可贵的是复杂度十分明确.\((O(nm+m))\)
该算法过程如下:
1. 进行n次松弛(对每条边, n为节点数).
2. 扫一遍边, 若还有可松弛的边说明有负环, 否则没有.
细节见代码.
- 代码 -
dfs
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int N = 500 + 5;
const int M = 5000 + 500;
int VIS[N], DIS[N];
int HD[N], NXT[M], TO[M], V[M];
int cas, n, m, w, sz;
bool flag;
void add(int frm, int to, int v) {
TO[++sz] = to; V[sz] = v; NXT[sz] = HD[frm]; HD[frm] = sz;
}
void init() {
sz = 0;
flag = false;
memset(HD, 0, sizeof (HD));
}
void spfa_dfs(int x) {
VIS[x] = 1;
for (int i = HD[x]; i; i = NXT[i]) {
int v = TO[i];
if (flag) break;
if (DIS[v] > DIS[x] + V[i]) {
if (VIS[v]) {
flag = true;
printf("YES\n");
break;
}
DIS[v] = DIS[x] + V[i];
spfa_dfs(v);
}
}
VIS[x] = 0;
}
int main() {
scanf("%d", &cas);
for (int cass = 1; cass <= cas; ++ cass) {
init();
int x, y, v;
scanf("%d%d%d", &n, &m, &w);
for (int i = 1; i <= m; ++i) {
scanf("%d%d%d", &x, &y, &v);
add(x, y, v);
add(y, x, v);
}
for (int i = 1; i <= w; ++i) {
scanf("%d%d%d", &x, &y, &v);
add(x, y, -v);
}
for (int i = 1; i <= n; ++i) {
if (flag) break;
memset(DIS, 0x3f, sizeof (DIS));
DIS[x] = 0;
spfa_dfs(i);
}
if (!flag) printf("NO\n");
}
return 0;
}
bellmanford
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int N = 500 + 5;
const int M = 5000 + 500;
int VIS[N], DIS[N];
int FRM[M], TO[M], V[M];
int cas, n, m, w, sz;
bool flag;
void add(int frm, int to, int v) {
TO[++sz] = to; V[sz] = v; FRM[sz] = frm;
}
void init() {
sz = 0;
memset(DIS, 0x3f, sizeof (DIS));
}
int bellmanford(int x) {
DIS[x] = 0;
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= sz; ++j)
if (DIS[TO[j]] > DIS[FRM[j]] + V[j])
DIS[TO[j]] = DIS[FRM[j]] + V[j];
for (int j = 1; j <= sz; ++j)
if (DIS[TO[j]] > DIS[FRM[j]] + V[j])
return 0;
return 1;
}
int main() {
scanf("%d", &cas);
for (int cass = 1; cass <= cas; ++ cass) {
init();
int x, y, v;
scanf("%d%d%d", &n, &m, &w);
for (int i = 1; i <= m; ++i) {
scanf("%d%d%d", &x, &y, &v);
add(x, y, v);
add(y, x, v);
}
for (int i = 1; i <= w; ++i) {
scanf("%d%d%d", &x, &y, &v);
add(x, y, -v);
}
int ans = bellmanford(1);
if (ans) printf("NO\n");
else printf("YES\n");
}
return 0;
}
