【POJ】3255 Roadblocks(次短絡＋spfa)
4329 ワード
http://poj.org/problem?id=3255
ハンガリーとゲームをします.
しかし、致命的なバグを発見しました.
ハンガリーでは、dis 2はシングルifではなく、else ifではなく、dis 2とdis 1は相対的に独立しています.前の2つのifを変更した後、より良いショートがあります.
だから、ワイキオは水が多すぎて、水が過ぎました.
Description
Bessie has moved d to a ssmaall farm and sometimes enjoys returning to visit one of her best friinds.She dost wantto to get to her old home too quickly、because she she she e likes the sceneralalininingtheway.She thethethethethethethethethethethethethethethethethethethethethethethekkkhatststsshshshshshshshshshshshshshshshshshshshshsheeeeeeeeeshshshshshshshshshshshshshshshshshshshshshshshshshshsheeeeeeeeth.
The countryside consists of R(1≦R≦100,000)bidirection roads、each linking two of the N(1≦N≦5000)intersections、conventlynumberd 1.N.Bessie starts at intersection 1、and her friend(the stination.)
The second-sharest path may share road s with any of the shotest paths,and it may backtrack i.e.use the same road or intersetion mone.The second-shotest path is the shot partest path path th th th th partest(i.e.if two or more shot paths exist、the second-shotest path is the one whose length is longer than butのlonger thany other path)
Input
Line 1:Two space-separated integers:N and R
Lines 2.R+1:Each LINE contains three space-separated integers:A,B,and D that describe a road that connects intersects A and B and has length D(1≦D≦5000)
Output
Line 1:The length of the second shotest path between node 1 and node N
Sample Input
Two routes:1->>2->4(length 100+200=300)and 1->2->3->4(length 100+250+100=450)
ソurce
USACO 2006 November Gold
ハンガリーとゲームをします.
しかし、致命的なバグを発見しました.
ハンガリーでは、dis 2はシングルifではなく、else ifではなく、dis 2とdis 1は相対的に独立しています.前の2つのifを変更した後、より良いショートがあります.
だから、ワイキオは水が多すぎて、水が過ぎました.
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
const int N=5050;
const long long oo=~0ull>>2;
int m, n, vis[N], q[N], front, tail, ihead[N], cnt;
long long d[N], d2[N];
struct ED { int to, next; long long w; }e[200010];
inline void add(const int &u, const int &v, const int &w) {
e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; e[cnt].w=w;
}
long long spfa(const int &s, const int &t) {
for1(i, 0, t) d[i]=d2[i]=oo;
d[s]=front=tail=0; vis[s]=1; q[tail++]=s;
int u, v, w;
while(front!=tail) {
u=q[front++]; if(front==N) front=0; vis[u]=0;
for(int i=ihead[u]; i; i=e[i].next) {
v=e[i].to; w=e[i].w;
if(d[v]>d[u]+w) {
d2[v]=d[v]; d[v]=d[u]+w;
if(!vis[v]) { vis[v]=1; q[tail++]=v; if(tail==N) tail=0; }
}
else if(d2[v]>d[u]+w && d[v]<d[u]+w) {
d2[v]=d[u]+w;
if(!vis[v]) { vis[v]=1; q[tail++]=v; if(tail==N) tail=0; }
}
if(d2[v]>d2[u]+w) {
d2[v]=d2[u]+w;
if(!vis[v]) { vis[v]=1; q[tail++]=v; if(tail==N) tail=0; }
}
}
}
if(d2[t]!=oo) return d2[t];
return -1;
}
int main() {
read(n); read(m);
int x, y, z;
rep(i, m) {
read(x); read(y); read(z);
add(x, y, z); add(y, x, z);
}
printf("%lld", spfa(1, n));
return 0;
}
Description
Bessie has moved d to a ssmaall farm and sometimes enjoys returning to visit one of her best friinds.She dost wantto to get to her old home too quickly、because she she she e likes the sceneralalininingtheway.She thethethethethethethethethethethethethethethethethethethethethethethekkkhatststsshshshshshshshshshshshshshshshshshshshshsheeeeeeeeeshshshshshshshshshshshshshshshshshshshshshshshshshshsheeeeeeeeth.
The countryside consists of R(1≦R≦100,000)bidirection roads、each linking two of the N(1≦N≦5000)intersections、conventlynumberd 1.N.Bessie starts at intersection 1、and her friend(the stination.)
The second-sharest path may share road s with any of the shotest paths,and it may backtrack i.e.use the same road or intersetion mone.The second-shotest path is the shot partest path path th th th th partest(i.e.if two or more shot paths exist、the second-shotest path is the one whose length is longer than butのlonger thany other path)
Input
Line 1:Two space-separated integers:N and R
Lines 2.R+1:Each LINE contains three space-separated integers:A,B,and D that describe a road that connects intersects A and B and has length D(1≦D≦5000)
Output
Line 1:The length of the second shotest path between node 1 and node N
Sample Input
4 4
1 2 100
2 4 200
2 3 250
3 4 100450Two routes:1->>2->4(length 100+200=300)and 1->2->3->4(length 100+250+100=450)
ソurce
USACO 2006 November Gold