1018.Public Bike Management (30)

1018.Public Bike Management (30)

pat-al-1018

2017-02-02

题意：找出耗时最短（单源最短路径）且send和back最少的路径

dijkstra加dfs

dfs中用了回溯法框架（其实我只学过回溯法，之前没有真的写过dfs……）

关于send和back的计算稍微有点难理解，在草稿纸上写写

对前驱节点使用了vector< int > pre来保存，因为可能不止一个，没法用数组

/**
* pat-al-1018
* 2017-02-01
* Cpp version
* Author: fengLian_s
*/
#include<cstdio>
#include<vector>
#include<cstring>
#define MAX 501
#define INF 0x3f3f3f3f
using namespace std;
vector<int> pre[MAX];
vector<int> path, tmpPath;
int D_valueOfBike[MAX];
int minSend = INF, minBack = INF;
void dfs(int s)
{
if(s == 0)
{
tmpPath.push_back(s);
int send = 0, back = 0;
for(int i = tmpPath.size()-1;i >= 0;i--)
{
int id = tmpPath[i];
if(D_valueOfBike[id] > 0)
back += D_valueOfBike[id];
else
{
if(back > (0 - D_valueOfBike[id]))
back += D_valueOfBike[id];
else
{
send += (0 - D_valueOfBike[id] - back);
back = 0;
}
}
}
if(send < minSend)
{
minSend = send;
minBack = back;
path = tmpPath;
}
else if(send == minSend && back < minBack)
{
minBack = back;
path = tmpPath;
}
tmpPath.pop_back();
return;
}
else//回溯法
{
tmpPath.push_back(s);
for(int i = 0;i < pre[s].size();i++)
dfs(pre[s][i]);
tmpPath.pop_back();
}
}
int main()
{
freopen("in.txt", "r", stdin);
int maxC, n, s, m;
int e[MAX][MAX], dist[MAX], visited[MAX];
scanf("%d %d %d %d", &maxC, &n, &s, &m);
for(int i = 1;i <= n;i++)
{
scanf("%d", &D_valueOfBike[i]);
D_valueOfBike[i] -= maxC/2;
}
memset(e, 0x3f, sizeof(e));
memset(dist, 0x3f, sizeof(dist));
memset(visited, 0, sizeof(visited));
for(int i = 0;i < m;i++)
{
int tmpS1, tmpS2, tmpTime;
scanf("%d %d %d", &tmpS1, &tmpS2, &tmpTime);
e[tmpS1][tmpS2] = tmpTime;
e[tmpS2][tmpS1] = tmpTime;
}
//dijkstra:
//printf("hello, dijkstra\n");
dist[0] = 0;
while(1)
{
int u, minD = INF;
for(int i = 0;i <= n;i++)
{
if(visited[i] == 0 && dist[i] < minD)
{
minD = dist[i];
u = i;
}
}
visited[u] = 1;//别忘记写了
if(minD == INF)
break;
for(int v = 0;v <= n;v++)
{
if(visited[v] == 0 && e[u][v] < INF)
{
//printf("e[%d][%d] = %d\n", u, v, e[u][v]);
if(dist[v] > dist[u] + e[u][v])
{
dist[v] = dist[u] + e[u][v];
pre[v].clear();
pre[v].push_back(u);
}
else if(dist[v] == dist[u] + e[u][v])
{
pre[v].push_back(u);
}
}
}
}
//dfs:
dfs(s);
//output:
//test:
// for(int i = 0;i < pre[s].size();i++)
// {
//   printf("%d\n", pre[s][i]);
// }
//test end
printf("%d 0", minSend);
for(int i = path.size()-2;i >= 0;i--)
{
printf("->%d", path[i]);
}
printf(" %d\n", minBack);
}

-FIN-