POJ 2197 Jill's Tour Paths (DFS)

题意：给定n个城市及其之间的距离，以及距离限制limit 初始点start 结束点end，要求求出从start->end的所有不大于limit的路径并输出，而且是按字典序输出。

思路：很明显DFS+记录路径，由于要求字典序输出，那么建立前向星表的时候就要注意了，搜索子节点的顺序一定是从小到大的！另外就是输出格式上注意一下，格式很huang很baoli！

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <memory.h>
#include <cmath>
#include <bitset>
#include <queue>
#include <vector>
using namespace std;

const int BORDER = (1<<20)-1;
const int MAXSIZE = 37;
const int MAXN = 30;
const int INF = 0x4ffffff;

#define CLR(x,y) memset(x,y,sizeof(x))
#define ADD(x) x=((x+1)&BORDER)
#define IN(x) scanf("%d",&x)
#define OUT(x) printf("%d\n",x)
#define MIN(m,v) (m)<(v)?(m):(v)
#define MAX(m,v) (m)>(v)?(m):(v)
#define ABS(x) ((x)>0?(x):-(x))

int n,m,ans;
int start,end,limit;
int index,net[MAXN];
int g[MAXN][MAXN];
bool visit[MAXN];
int route[MAXN];
int deep;
int case_id;

typedef struct{
    int v,val,next;
}EDGE;
typedef struct{
    int len,num;
    int city[MAXN];
}NODE;
EDGE edge[MAXN*MAXN];
NODE node[MAXN*MAXN*MAXN*MAXN];

int init()
{
    CLR(net,-1);
    CLR(visit,0);
    CLR(g,0);
    index = 0;
    ans = 0;
    deep = 0;
    return 0;
}
void add_edge(const int& u,const int& v,const int& val)
{
    edge[index].next = net[u];
    edge[index].v = v;
    edge[index].val = val;
    net[u] = index;
    ++index;
}
bool cmp(const NODE& a,const NODE& b)
{
    return a.len < b.len;
}
/*
{
    if(a.len != b.len)
        return a.len < b.len;
    int t = MAX(a.num,b.num);
    for(int i = 1; i <= t; ++i)
        if(a.city[i] != b.city[i])
            return a.city[i] < b.city[i];
    return true;
}
*/
int input()
{
    int i,j,tmp;
    int a,b;
    scanf("%d",&m);
    for(i = 1; i <= m; ++i)
    {
        scanf("%d%d%d",&a,&b,&tmp);
        g[a][b] = g[b][a] = tmp;
    }
    scanf("%d%d",&start,&end);
    scanf("%d",&limit);
    return 0;
}
int make_graph()
{
    int i,j,tmp;
    for(i = 1; i <= n; ++i)
        for(j = n; j >= 1; --j)
            if(g[i][j])
                add_edge(i,j,g[i][j]);
    return 0;
}
void dfs(const int& u,const int& dist)
{
    if(dist > limit)
        return ;
    route[deep] = u;
    ++deep;
    if(u == end)
    {
        memcpy(node[ans].city,route,deep*sizeof(route[0]));
        node[ans].len = dist;
        node[ans].num = deep;
        ++ans;
        --deep;
        return ;
    }
    for(int i = net[u] ; i != -1; i = edge[i].next)
        if(!visit[edge[i].v])
        {
            visit[edge[i].v] = true;
            dfs(edge[i].v,dist+edge[i].val);
            visit[edge[i].v] = false;
        }
    --deep;
}
int output()
{
    int i,j,tmp;
    printf("Case %d:\n",case_id);
    if(ans == 0)
    {
        printf(" NO ACCEPTABLE TOURS\n\n");
        return 0;
    }
    for(i = 0; i < ans; ++i)
    {
        printf(" %d:",node[i].len);
        for(j = 0; j < node[i].num; ++j)
            printf(" %d",node[i].city[j]);
        printf("\n");
    }
    printf("\n");
    return 0;
}
int work()
{
    make_graph();
    visit[start] = true;
    dfs(start,0);
    sort(node,node+ans,cmp);
    return 0;
}
int main()
{
    case_id = 1;
    while(scanf("%d",&n) != EOF)
    {
        if(n == -1)
            break;
        init();
        input();
        work();
        output();
        ++case_id;
    }
    return 0;
}