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poj 3255 Roadblocks
Roadblocks
Time Limit: 2000MS |
| Memory Limit: 65536K |
Total Submissions: 13216 |
| Accepted: 4660 |
Description
Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.
The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersectionN.
The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any 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 intersections A and B and has length D (1 ≤ D ≤ 5000)
Output
Line 1: The length of the second shortest path between node 1 and node N
Sample Input
4 4
1 2 100
2 4 200
2 3 250
3 4 100
Sample Output
450
Hint
Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)
Source
USACO 2006 November Gold
题意：
给出n个点，m条双向边，求严格次短路。
AC代码：
#include<cstdio>#include<cstring>#include<queue>#define R registerusing namespace std;inline int read(){ R int x=0;bool f=1; R char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=0;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=(x<<3)+(x<<1)+ch-‘0‘;ch=getchar();} return f?x:-x;}const int N=1e5+10;struct node{ int u,v,w,next;}e[N<<1];int n,m,tot,head[N],dis1[N],dis2[N];bool vis[N];void add(int x,int y,int z){ e[++tot].u=x; e[tot].v=y; e[tot].w=z; e[tot].next=head[x]; head[x]=tot;}void spfa1(int S){ queue<int>q; memset(vis,0,sizeof vis); memset(dis1,127/3,sizeof dis1); q.push(S); dis1[S]=0;vis[S]=1; while(!q.empty()){ int x=q.front();q.pop(); vis[x]=0; for(int i=head[x];i;i=e[i].next){ int v=e[i].v,w=e[i].w; if(dis1[v]>dis1[x]+w){ dis1[v]=dis1[x]+w; if(!vis[v]){ vis[v]=1; q.push(v); } } } }}void spfa2(int S){ queue<int>q; memset(vis,0,sizeof vis); memset(dis2,127/3,sizeof dis2); q.push(S); dis2[S]=0;vis[S]=1; while(!q.empty()){ int x=q.front();q.pop(); vis[x]=0; for(int i=head[x];i;i=e[i].next){ int v=e[i].v,w=e[i].w; if(dis2[v]>dis2[x]+w){ dis2[v]=dis2[x]+w; if(!vis[v]){ vis[v]=1; q.push(v); } } } }}void Cl(){ tot=0; memset(e,0,sizeof e); memset(head,0,sizeof head);}void work(){ Cl(); for(int i=1,x,y,z;i<=m;i++){ x=read();y=read();z=read(); add(x,y,z); add(y,x,z); } spfa1(1); spfa2(n); int shortest=dis1[n],shorter=0x7fffffff; for(int i=1;i<=m*2;i++){ int len=dis1[e[i].u]+dis2[e[i].v]+e[i].w; if(len>shortest&&len<shorter) shorter=len; } printf("%d\n",shorter);}int main(){ while(scanf("%d%d",&n,&m)==2) work(); return 0;}
poj 3255 Roadblocks