洛谷 P1119 灾后重建

题目大意：

传送门

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解题思路：

看到数据很明显的 $Floyd​$

但是问题来了，如果每次询问都跑一遍 $Floyd$时间复杂度是： $O(Qn^3)$，会 $T$飞

我们发现询问的时间是递增的，那么我们可以在 $Floyd$找中转站的循环，也就是第一层循环中判断是否能在中转站修好之后走

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$Accepted\ code:$

#include<cstdio>
#include<algorithm>

using namespace std;

const int inf = 1e9;

int n, m, Q, x, y, w;
int t[205], f[205][205];

int main() {
    scanf("%d %d", &n, &m);
    for (int i = 0; i <= n+1; ++i)
        for (int j = 0; j <= n+1; ++j)
            f[i][j] = inf;
    for (int i = 1; i <= n; ++i)
        scanf("%d", &t[i]);
    t[n+1] = inf; 
    for (int i = 1; i <= m; ++i)
        scanf("%d %d %d", &x, &y, &w),
        f[x+1][y+1] = f[y+1][x+1] = min(f[x+1][y+1], w);
    scanf("%d", &Q); int k = 1;
    while (Q--) {
        scanf("%d %d %d", &x, &y, &w); ++x; ++y;
        if (t[x] > w || t[y] > w) { puts("-1"); continue; }
        while (t[k] <= w) {
            for (int i = 1; i <= n; ++i)
                for (int j = 1; j <= n; ++j)
                    f[i][j] = min(f[i][j], f[i][k]+f[k][j]);
            ++k;
        }
        printf("%d\n", f[x][y] < inf ? f[x][y] : -1);
    }
}