1018 Public Bike Management (30)（30 分）

时间限制400 ms
内存限制65536 kB
代码长度限制16000 B

There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it to any other stations in the city.

The Public Bike Management Center (PBMC) keeps monitoring the real-time capacity of all the stations. A station is said to be in perfect condition if it is exactly half-full. If a station is full or empty, PBMC will collect or send bikes to adjust the condition of that station to perfect. And more, all the stations on the way will be adjusted as well.

When a problem station is reported, PBMC will always choose the shortest path to reach that station. If there are more than one shortest path, the one that requires the least number of bikes sent from PBMC will be chosen.

Figure 1
Figure 1 illustrates an example. The stations are represented by vertices and the roads correspond to the edges. The number on an edge is the time taken to reach one end station from another. The number written inside a vertex S is the current number of bikes stored at S. Given that the maximum capacity of each station is 10. To solve the problem at S3, we have 2 different shortest paths:

1. PBMC -> S1 -> S3. In this case, 4 bikes must be sent from PBMC, because we can collect 1 bike from S1 and then take 5 bikes to S3, so that both stations will be in perfect conditions.

2. PBMC -> S2 -> S3. This path requires the same time as path 1, but only 3 bikes sent from PBMC and hence is the one that will be chosen.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 numbers: Cmax (<= 100), always an even number, is the maximum capacity of each station; N (<= 500), the total number of stations; Sp, the index of the problem station (the stations are numbered from 1 to N, and PBMC is represented by the vertex 0); and M, the number of roads. The second line contains N non-negative numbers Ci (i=1,…N) where each Ci is the current number of bikes at Si respectively. Then M lines follow, each contains 3 numbers: Si, Sj, and Tij which describe the time Tij taken to move betwen stations Si and Sj. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print your results in one line. First output the number of bikes that PBMC must send. Then after one space, output the path in the format: 0->S1->…->Sp. Finally after another space, output the number of bikes that we must take back to PBMC after the condition of Sp is adjusted to perfect.

Note that if such a path is not unique, output the one that requires minimum number of bikes that we must take back to PBMC. The judge’s data guarantee that such a path is unique.

Sample Input:
10 3 3 5
6 7 0
0 1 1
0 2 1
0 3 3
1 3 1
2 3 1
Sample Output:
3 0->2->3 0

上去一看最短路，顺便附带空缺（need）和多余（nneed）的个数记录，然后有两个点没过，抠了半天，他的要求是这条路径中，如果有某个点缺车，而前面总的多余的话，可以填空缺，如果前面不多，这里的空缺就只能由总站提供，我一直以为他是在一条路径走个来回，某个点多余的可以同时填补前面或者后面的空缺，再者，要求出所有的最短路径，然后找出最优的。

不能只用最短路去满足所有要求，单纯对于一个不是查询点的点，假如按照最优去更新他，那么可能是不满足条件的，从他到目标点的路上可能有过多的空缺，需要填补，这个时候就希望在这个点之前有足够多的多余，可是总的要求是在到某点路径同样短且空缺相同的情况下，多余的车辆尽量少，就与后方高需求矛盾了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
#define inf 0x3f3f3f3f
int c;///每个车站最大容量
int n;///车站的总个数
int p;///要求车站的编号
int m;///道路的条数
int current[501];///各车站当前车辆数
int road[501][501];///两车站距离
int dis[501];///0到i路径的长度
int vis[501];///标记访问过该车站没有
vector<int> pathfrom[501];///记录最短路径来源
vector<int> temppath;///临时路径记录
vector<int> path;///记录最佳路径
int perfect;///最佳状态
int minneed = inf,minnneed = inf;
int u,v,w;
void dfs(int ve)
{
    if(!ve)
    {
        int need = 0,nneed = 0;
        for(int i = temppath.size() - 2;i >= 0;i --)
        {
            if(current[temppath[i]] >= 0)nneed += current[temppath[i]];
            else
            {
                if(nneed + current[temppath[i]] >= 0)
                {
                    nneed += current[temppath[i]];
                }
                else
                {
                    need -= (nneed + current[temppath[i]]);
                    nneed = 0;
                }
            }
        }
        if(need < minneed)minneed = need,minnneed = nneed,path = temppath;
        else if(need == minneed && nneed < minnneed)minnneed = nneed,path = temppath;
        return;
    }
    for(int i = 0;i < pathfrom[ve].size();i ++)
    {
        int d = pathfrom[ve][i];
        temppath.push_back(d);
        dfs(d);
        temppath.pop_back();
    }
}
int main()
{
    scanf("%d%d%d%d",&c,&n,&p,&m);
    c /= 2;
    perfect -= c * n;
    for(int i = 1;i <= n;i ++)
    {
        scanf("%d",&current[i]);
        current[i] -= c;
    }
    for(int i = 0;i <= n;i ++)
    {
        for(int j = 0;j <= n;j ++)
        {
            road[i][j] = inf;
        }
        dis[i] = inf;
        road[i][i] = 0;
    }
    for(int i = 0;i < m;i ++)
    {
        scanf("%d%d%d",&u,&v,&w);
        road[u][v] = road[v][u] = w;
    }
    int t;///最短距离的结点
    int mi;///最短距离
    dis[0] = 0;
    while(1)
    {
        t = -1;
        mi = inf;
        for(int i = 0;i <= n;i ++)
        {
            if(!vis[i] && mi > dis[i])
            {
                t = i;
                mi = dis[i];
            }
        }
        if(t == -1)break;
        vis[t] = 1;
        for(int i = 0;i <= n;i ++)
        {
            if(vis[i] || road[t][i] == inf)continue;
            int d = dis[t] + road[t][i];
            if(d < dis[i])
            {
                pathfrom[i].clear();
                dis[i] = d;
                pathfrom[i].push_back(t);
            }
            else if(d == dis[i])
            {
                pathfrom[i].push_back(t);
            }
        }
    }
    temppath.push_back(p);
    dfs(p);
    printf("%d 0",minneed);
    for(int i = path.size() - 2;i >= 0;i --)
        printf("->%d",path[i]);
    printf(" %d",minnneed);
}

错误代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define inf 0x3f3f3f3f
int c;///每个车站最大容量
int n;///车站的总个数
int p;///要求车站的编号
int m;///道路的条数
int current[501];///各车站当前车辆数
int road[501][501];///两车站距离
int dis[501];///0到i路径的长度
int num[501];///0到i车辆的总数
int vis[501];///标记访问过该车站没有
int path[501];///路径
int spath[501];///最短路径
int pa;///spath下标
int u,v,w;
void getpath(int s)
{
    if(path[s])getpath(path[s]);
    spath[pa ++] = s;
}
int main()
{
    scanf("%d%d%d%d",&c,&n,&p,&m);
    c /= 2;
    for(int i = 1;i <= n;i ++)
    {
        scanf("%d",&current[i]);
        current[i] -= c;
    }
    for(int i = 0;i <= n;i ++)
    {
        for(int j = 0;j <= n;j ++)
        {
            road[i][j] = inf;
        }
        road[i][i] = 0;
    }
    for(int i = 0;i < m;i ++)
    {
        scanf("%d%d%d",&u,&v,&w);
        road[u][v] = road[v][u] = w;
    }
    int t;///最短距离的结点
    int mi;///最短距离
    for(int i = 1;i <= n;i ++)
    {
        dis[i] = road[0][i];
        if(dis[i] != inf)
        {
            num[i] = current[i];
            path[i] = 0;
        }
    }
    while(1)
    {
        mi = inf;
        for(int i = 1;i <= n;i ++)
        {
            if(!vis[i] && mi > dis[i])
            {
                mi = dis[t = i];
            }
        }
        if(mi == inf)break;
        vis[t] = 1;
        for(int i = 1;i <= n;i ++)
        {
            if(vis[i])continue;
            int d = dis[t] + road[t][i];
            if(d < dis[i])
            {
                dis[i] = d;
                num[i] = num[t] + current[i];
                path[i] = t;
            }
            else if(d == dis[i])
            {
                int change = num[t] + current[i];
                if(num[i] < 0 && change > num[i] || num[i] >= 0 && change >= 0 && num[i] > change)
                {
                    num[i] = change;
                    path[i] = t;
                }
            }
        }
    }
    getpath(p);
    printf("%d 0",num[p] < 0 ? -num[p] : 0);
    for(int i = 0;i < pa;i ++)
        printf("->%d",spath[i]);
    printf(" %d",num[p] <= 0 ? 0 : num[p]);
}