POJ2449题解

• 先讲一个为了少打一些代码而滥用继承终于接受慘痛教训的故事。

#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
const int oo = 1000000000, nil = 0;
int N, M, S, T, K, times[1005];
int u[200010], v[200010], w[200010], nxt[200010], pnt[1005], e;
int d[1005];
bool vis[1005], other[1005];
class node
{
    public:
        int n, dis;
        node(int n = 0, int dis = 0) :n(n), dis(dis) {}
    protected:
        virtual bool operator < (const node& b) const
        {
            return dis > b.dis;
        }
};
struct astar : public node
{
    astar(int xn = 0, int xdis = 0)
    {
        n = xn; dis = xdis;
    }
    bool operator < (const astar& b) const
    {
        return dis + d[n] > b.dis + d[b.n];
    }
};
void addedge(int a, int b, int c, bool d)
{
    u[++e] = a; v[e] = b; w[e] = c;
    nxt[e] = pnt[a]; pnt[a] = e; other[e] = d;
}
void init()
{
    int a, b, c;
    scanf("%d%d", &N, &M);
    for(int i = 1; i <= M; ++i)
    {
        scanf("%d%d%d", &a, &b, &c);
        addedge(a, b, c, false);
        addedge(b, a, c, true);
    }
    scanf("%d%d%d", &S, &T, &K);
    if(S == T)
    {
        ++K;
    }
}
void getSSSP(int s)
{
    memset(d, 0x3f, sizeof(d));
    memset(vis, 0, sizeof(vis));
    priority_queue <node> Q;
    d[s] = 0;
    Q.push(node(s, 0));
    while(!Q.empty())
    {
        node t = Q.top();
        Q.pop();
        vis[t.n] = true;
        for(int j = pnt[t.n]; j != nil; j = nxt[j])
        {
            if((!vis[v[j]]) && (d[v[j]] > t.dis + w[j]) && (other[j]))
            {
                d[v[j]] = t.dis + w[j];
                Q.push(node(v[j], d[v[j]]));
            }
        }
    }
}
void work()
{
    getSSSP(T);
    memset(times, 0, sizeof(times));
    priority_queue <astar> Q;
    Q.push(astar(S, 0));
    while(!Q.empty())
    {
        astar t = Q.top();
        Q.pop();
        ++times[t.n];
        if((t.n == T) && (times[t.n] == K))
        {
            printf("%d\n", t.dis);
            return;
        }
        for(int j = pnt[t.n]; j != nil; j = nxt[j])
        {
            if(!other[j])
            {
                Q.push(astar(v[j], t.dis + w[j]));
            }
        }
    }
    puts("-1");
}
int main()
{
    init();
    work();
    return 0;
}

然后由于node里的小于号被protected了。stl的priority_queue无法调用之而编译失败。

假设去掉protected。node里的小于号覆盖不掉，又会使astar中的小于号没用导致WA。TUT……

• 题目来源

id=2449">http://poj.org/problem?id=2449

• 题目大意
  
  求一张有向图从S到T的第K短路（边权均为正）。

• 题解
  
  首先在反向图中最短路出每一个点到终点的距离，然后利用A*的思想，从起点開始可反复地搜索，估价函数为{当前走过的距离+该点到终点的最短路长度}。当终点T被第K次从堆（优先队列）中取出时，输出答案。

• 注意细节
  
  S == T时要把K加上1；
  
  某结点出队次数超过K时，能够不再考虑。

• Code

#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
const int oo = 1000000000, nil = 0;
int N, M, S, T, K, times[1005];
int u[200010], v[200010], w[200010], nxt[200010], pnt[1005], e;
int d[1005];
bool vis[1005], other[200010];
struct node
{
    int n, dis;
    node(int n = 0, int dis = 0) :n(n), dis(dis) {}
    bool operator < (const node& b) const
    {
        return dis > b.dis;
    }
};
struct astar
{
    int n, dis;
    astar(int n = 0, int dis = 0) :n(n), dis(dis) {}
    bool operator < (const astar& b) const
    {
        return dis + d[n] > b.dis + d[b.n];
    }
};
void addedge(int a, int b, int c, bool d)
{
    u[++e] = a; v[e] = b; w[e] = c;
    nxt[e] = pnt[a]; pnt[a] = e; other[e] = d;
}
void init()
{
    int a, b, c;
    scanf("%d%d", &N, &M);
    for(int i = 1; i <= M; ++i)
    {
        scanf("%d%d%d", &a, &b, &c);
        addedge(a, b, c, false);
        addedge(b, a, c, true);
    }
    scanf("%d%d%d", &S, &T, &K);
    if(S == T)
    {
        ++K;
    }
}
void getSSSP()
{
    memset(d, 0x3f, sizeof(d));
    memset(vis, 0, sizeof(vis));
    priority_queue <node> Q;
    d[T] = 0;
    Q.push(node(T, 0));
    while(!Q.empty())
    {
        node t = Q.top();
        Q.pop();
        vis[t.n] = true;
        for(int j = pnt[t.n]; j != nil; j = nxt[j])
        {
            if((!vis[v[j]]) && (d[v[j]] > t.dis + w[j]) && (other[j]))
            {
                d[v[j]] = t.dis + w[j];
                Q.push(node(v[j], d[v[j]]));
            }
        }
    }
}
void work()
{
    getSSSP();
    memset(times, 0, sizeof(times));
    priority_queue <astar> Q;
    Q.push(astar(S, 0));
    while(!Q.empty())
    {
        astar t = Q.top();
        Q.pop();
        ++times[t.n];
        if((t.n == T) && (times[t.n] == K))
        {
            printf("%d\n", t.dis);
            return;
        }
        if(times[t.n] > K)
        {
            continue;
        }
        for(int j = pnt[t.n]; j != nil; j = nxt[j])
        {
            if(!other[j])
            {
                Q.push(astar(v[j], t.dis + w[j]));
            }
        }
    }
    puts("-1");
}
int main()
{
    init();
    work();
    return 0;
}

• 再讲一个悲伤的故事：这段代码我一開始把other数组开成了点数，然后RE了4次，另一次忘了删去while(1)……所以一定要养成提前定义常量和认真静态查错的好习惯。