災后重建

luogu 1119

題目大意

給你一個圖，每個點在一定時間后才能被通過，問在某個時間，兩個點之間的最短路

輸入樣例

4 5 1 2 3 4 0 2 1 2 3 1 3 1 2 2 1 4 0 3 5 4 2 0 2 0 1 2 0 1 3 0 1 4

輸出樣例

-1 -1 5 4

數據范圍

$N?200，M?\frac{N\times (N?1)}{2}，Q?50000$

解題思路

題目保證每個點的通行時間是單調遞增的
那么可以離線處理
當連上一個點后再以這個點為中間點跑$Floyd$
輸出的時候判斷一下兩個點的通行時間是否小于查詢時間即可

代碼

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ll long long #define N 210 #define Q 50010 using namespace std; ll n, m, x, y, z, q, qt, t[N], a[N][N]; struct node {ll x, y, t, ans; }ask[Q]; ll read() {ll ds = 0, fs = 1;char c = getchar();while(c < '0' || c > '9') {if (c == '-') fs = -1; c = getchar();}while('0' <= c && c <= '9') ds = (ds<<1) + (ds<<3) + c - 48, c = getchar();return ds * fs; } void writ(ll x) {if (x / 10) writ(x / 10);putchar(x % 10 + 48); } void write(ll x) {if (x < 0) putchar('-'), x = -x;writ(x);putchar(10);return; } int main() {memset(a, 127/3, sizeof(a));n = read();m = read();for (ll i = 0; i < n; ++i)t[i] = read();for (ll i = 1; i <= m; ++i){x = read();y = read();z = read();a[x][y] = min(a[x][y], z);a[y][x] = min(a[y][x], z);}q = read();for (ll i = 1; i <= q; ++i){ask[i].x = read();ask[i].y = read();ask[i].t = read();}qt = 1;for (ll k = 0; k < n; ++k){while(ask[qt].t < t[k] && qt <= q)//離線處理{ask[qt].ans = a[ask[qt].x][ask[qt].y];qt++;}for (ll i = 0; i < n; ++i)for (ll j = 0; j < n; ++j)a[i][j] = min(a[i][j], a[i][k] + a[k][j]);//Floyd}while(qt <= q){ask[qt].ans = a[ask[qt].x][ask[qt].y];qt++;}for (ll i = 1; i <= q; ++i)if (ask[i].t < t[ask[i].x] || ask[i].t < t[ask[i].y] || ask[i].ans > 2000000) write(-1);//判斷是否可以通行else write(ask[i].ans);return 0; }

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