灾后重建 Floyd-CFANZ编程社区

题目背景

灾后重建 Floyd_#define

题目描述

给出灾后重建 Floyd_#include_02

输入输出格式

输入格式：

第一行包含两个正整数灾后重建 Floyd_#define_03

第二行包含灾后重建 Floyd_数据_04

接下来灾后重建 Floyd_#include_05

接下来一行也就是灾后重建 Floyd_#define_06

接下来灾后重建 Floyd_数据_07

输出格式：

共灾后重建 Floyd_#define_08

输入输出样例

输入样例#1：

复制

输出样例#1： 复制

-1
-1
5
4

说明

对于灾后重建 Floyd_#include_09

对于灾后重建 Floyd_#include_10

对于灾后重建 Floyd_#define_11

对于灾后重建 Floyd_#define_12

$知识点不难，重要是idea;$

$由于保证 t 是严格递增的，那么我们每次更新就行了；$

$floyd;$

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair pii;
inline ll rd() {
  ll x = 0;
  char c = getchar();
  bool f = false;
  while (!isdigit(c)) {
    if (c == '-') f = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f ? -x : x;
}

ll gcd(ll a, ll b) {
  return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }


/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
  if (!b) {
    x = 1; y = 0; return a;
  }
  ans = exgcd(b, a%b, x, y);
  ll t = x; x = y; y = t - a / b * y;
  return ans;
}
*/

int dis[300][300];
int n, m;
int fixt[maxn];

void floyd(int k) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (dis[i][j] > dis[i][k] + dis[k][j])
        dis[i][j] = dis[j][i] = dis[i][k] + dis[k][j];
    }
  }
}

int main() {
  //ios::sync_with_stdio(0);
  rdint(n); rdint(m);
  for (int i = 0; i < n; i++)rdint(fixt[i]);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++)dis[i][j] = dis[j][i] = 1e9;
  }
  for (int i = 0; i <= n; i++) {
    dis[i][i] = 0;
  }
  for (int i = 0; i < m; i++) {
    int u, v, w; rdint(u); rdint(v); rdint(w);
    dis[u][v] = dis[v][u] = w;
  }
  int cur = 0;
  int q; cin >> q;
  while (q--) {
    int x, y, t; rdint(x); rdint(y); rdint(t);
    while (cur < n&&fixt[cur] <= t) {
      floyd(cur); cur++;
    }
    if (fixt[x] > t || fixt[y] > t) {
      cout << -1 << endl;
    }
    else {
      if (dis[x][y] == 1e9)cout << -1 << endl;
      else cout << dis[x][y] << endl;
    }
  }
  return 0;
}

EPFL - Fighting