UVa10269 Adventure of Super Mario

不错的最短路题.一开始没想到如何处理 "Mario NEVER super runs through a Castle" 这个条件,然后大力分类讨论.

在写了将近70行的状态扩展后,我放弃了.

实际上,这题没有这么麻烦.这个条件限制只用在floyd预处理的时候就可以解决了.

具体来说, 在floyd的时候,限制中转点为乡村

        for(int k = 1; k <= A; k++)
            for(int i = 1; i <= (A + B); i++)
                for(int j = 1; j <= (A + B); j++)

 根据算法的性质, 这样所有路径除起点, 终点之外的点就都在乡村了.

用了这个非常妙的做法以后, 就不难写出状态的扩展了.

这里用的是dijkstra

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 1e2 + 20;
const int INF = 0x3f3f3f3f;

int A, B, M, L, K;
int d[MAXN][MAXN], f[MAXN][12];
bool vis[MAXN][12];

struct edge
{
    int to, cost;
};vector<edge> g[MAXN];
struct state
{
    int pos, time, k;
    state(int u = 0, int t = 0, int k = 0) :
    pos(u), time(t), k(k) {}
    bool operator >(const state &rhs) const{
        return time > rhs.time;
    }
};

void init()
{
    memset(d, 0x3f, sizeof(d));
    memset(f, 0x3f, sizeof(f));
    memset(vis, false, sizeof(vis));
    for(int i = 1; i <= (A + B); i++) g[i].clear();
}

void floyd()
{
    for(int i = 1; i <= (A + B); i++) d[i][i] = 0;
        for(int k = 1; k <= A; k++)
            for(int i = 1; i <= (A + B); i++)
                for(int j = 1; j <= (A + B); j++)
                    if(d[i][k] != INF && d[k][j] != INF)
                        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}

inline bool tension(const int &st, int &lg)
{
    return st < lg ? (lg = st, true) : false;
}

int bfs()
{
    priority_queue<state, vector<state>, greater<state> > q;
    f[(A + B)][0] = 0;
    q.push(state((A + B), 0, 0));

    while(!q.empty())
    {
        state u = q.top(); q.pop();
        if(u.pos == 1) return u.time;
        if(vis[u.pos][u.k]) continue;
        vis[u.pos][u.k] = true;
        for(int i = 0; i < (int) g[u.pos].size(); i++)
        {
            edge &e = g[u.pos][i];
            if(tension(u.time + e.cost, f[e.to][u.k]))
                q.push(state(e.to, f[e.to][u.k], u.k));
        }

        if(u.k < K)
        for(int i = 1; i <= (A + B); i++)
            if(d[i][u.pos] <= L && tension(u.time + 0, f[i][u.k + 1]))
                q.push(state(i, f[i][u.k + 1], u.k + 1));
    }
    return rand();
}

int main()
{
    //freopen("10269.in", "r", stdin);
    //freopen("10269.out", "w", stdout);
    int T; cin>>T;
    while(T--)
    {
        cin>>A>>B>>M>>L>>K;init();
        for(int i = 1, u, v, c; i <= M; i++)
        {
            scanf("%d%d%d", &u, &v, &c);
            d[u][v] = d[v][u] = c;
            g[u].push_back((edge){v, c});
            g[v].push_back((edge){u, c});
        }
        floyd();
        printf("%d
", bfs());
    }
    return 0;
}