POJ 3411 Paid Roads

Description

A network of m roads connects N cities (numbered from 1 to N). There may be more than one road connecting one city with another. Some of the roads are paid. There are two ways to pay for travel on a paid road i from city ai to city bi:

in advance, in a city ci (which may or may not be the same as ai);

after the travel, in the city bi.

The payment is Pi in the first case and Ri in the second case.

Write a program to find a minimal-cost route from the city 1 to the city N.

解题报告:

这个题花了很久不知道错在哪，后来发现对题目意思理解不是很透彻，仿佛他经过一个点Ci一次只能获得走i这条边的机会，然后可以在Ci一次性买多次，我写的spfa，状态表示为f[i][j] 表示在i这个点，可以经过的边的状态为j的最小费用,然后发现j不能表示为0/1,可以为2/3及以上，最后改写成深搜并改变状态即可

WA的SPFA:

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define RG register
#define il inline
#define iter iterator
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int N=500,M=1<<11;
int n,m,head
,num=0,nxt[N<<1],to[N<<1],dis[N<<1],id[N<<1];
int NXT[N<<1],To[N<<1],Head
,NUM=0,Pai[N<<1];
void addlink(int x,int y,int paid){
NXT[++NUM]=Head[x];To[NUM]=1<<(y-1);Pai[NUM]=paid;Head[x]=NUM;
}
void link(int x,int y,int z,int ids){
nxt[++num]=head[x];to[num]=y;
id[num]=1<<(ids-1);dis[num]=z;head[x]=num;
}
struct node{
int x,t;
node(){}
node(int _x,int _t){x=_x;t=_t;}
}q[1000005];
int mod=1000005,f
[M];bool vis
[M];
void spfa(){
int t=0,sum=1,x,k,u,kis,paid;
memset(f,127/3,sizeof(f));
q[1].x=1;q[1].t=0;f[1][0]=0;vis[1][0]=true;
while(t!=sum){
t++;if(t==mod)t=0;
x=q[t].x;k=q[t].t;
for(int i=head[x];i;i=nxt[i]){
u=to[i];
if(k&id[i]){
kis=k-id[i];paid=0;
}
else kis=k,paid=dis[i];
if(f[x][k]+paid<f[u][kis]){
f[u][kis]=f[x][k]+paid;
if(!vis[u][kis]){
vis[u][kis]=true;
sum++;if(sum==mod)sum=0;
q[sum]=node(u,kis);
}
}
}
for(int i=Head[x];i;i=NXT[i]){
u=To[i];if(u&k)continue;
kis=u|k;
if(f[x][k]+Pai[i]<f[x][kis]){
f[x][kis]=f[x][k]+Pai[i];
if(!vis[x][kis]){
vis[x][kis]=true;
sum++;if(sum==mod)sum=0;
q[sum]=node(x,kis);
}
}
}
vis[x][k]=false;
}
}
void work()
{
int x,y,c,p,r;
for(int i=1;i<=m;i++){
scanf("%d%d%d%d%d",&x,&y,&c,&p,&r);
link(x,y,r,i);addlink(c,i,p);
}
spfa();
int ans=2e8,lim=1<<m;
for(int i=0;i<lim;i++){
if(f
[i]<ans)ans=f
[i];
}
if(ans==2e8)puts("impossible");
else printf("%d\n",ans);
}

int main()
{
while(~scanf("%d%d",&n,&m))
work();
return 0;
}

深搜:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const int maxn = 15;
struct edge
{
int v,c,p,r,next;
}edges[maxn];
int head[maxn];
int vispath[maxn];
int n,m,e,ans;

void addedge(int u,int v,int c,int p,int r)
{
edges[e].v = v; edges[e].c = c; edges[e].p = p; edges[e].r = r;
edges[e].next = head[u];
head[u] = e++;
}

void dfs(int u,int cost)
{
if(cost > ans) return;
if(u == n)
{
if(cost < ans) ans = cost;
return;
}
for(int i=head[u];i!=-1;i=edges[i].next)
{
int v = edges[i].v;
if(vispath[v]<=3)
{
vispath[v]++;
if(vispath[edges[i].c]>0) dfs(v,cost+edges[i].p);
else dfs(v,cost+edges[i].r);
vispath[v]--;
}
}
}

int main()
{
freopen("pp.in","r",stdin);
int u,v,c,p,r;
e = 0;
memset(head,-1,sizeof(head));
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
scanf("%d%d%d%d%d",&u,&v,&c,&p,&r);
addedge(u,v,c,p,r);
}
ans = 99999999;
vispath[1] = 1;
dfs(1,0);
if(ans < 99999999) printf("%d\n",ans);
else printf("impossible\n");
return 0;
}