UVa 10269 Adventure of Super Mario (Floyd + DP + BFS)

题意：有A个村庄，B个城市，m条边，从起点到终点，找一条最短路径。但是，有一种工具可以使人不费力的移动L个长度，但始末点必须是城市或村庄。这种工具有k个，每个只能使用一次，并且在城市内部不可使用，但在村庄内部可使用。另外，在城市或村庄内部的时间不计。

析：先预处理出来使用工具能到达的距离，这个可以用Floyd 来解决，然后f[i][u] 表示到达 u 还剩下 i 次工具未用，然后用bfs就可以很简单的解决。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define pu push_up

#define pd push_down

#define cl clear()

#define all 1,n,1

#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 100 + 10;

const int maxm = 1e5 + 10;

const int mod = 50007;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, -1, 0, 1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

int dp[maxn][maxn];

int f[12][maxn];

int main(){

  int T;  cin >> T;

  while(T--){

    int A, B, L, K;

    scanf("%d %d %d %d %d", &A, &B, &m, &L, &K);

    ms(dp, INF);

    while(m--){

      int u, v, c;  scanf("%d %d %d", &u, &v, &c);

      dp[u][v] = dp[v][u] = min(c, dp[u][v]);

    }

    FOR(k, 1, A+1)  FOR(i, 1, A+B+1)  FOR(j, 1, A+B+1)

      dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);

    ms(f, INF);  f[K][A+B] = 0;

    queue<P> q;  q.push(P(K, A+B));

    while(!q.empty()){

      P p = q.front();  q.pop();

      int u = p.se, i = p.fi;

      for(int v = 1; v < A+B; ++v){

        if(dp[u][v] == INF || u == v)  continue;

        if(f[i][v] > f[i][u] + dp[u][v]){

          f[i][v] = f[i][u] + dp[u][v];

          q.push(P(i, v));

        }

        if(i && f[i-1][v] > f[i][u] && L >= dp[u][v]){

          f[i-1][v] = f[i][u];

          q.push(P(i-1, v));

        }

      }

    }

    int ans = INF;

    for(int i = 0; i <= K; ++i)  ans = min(ans, f[i][1]);

    printf("%d\n", ans);

  }

  return 0;

}