Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P1948 [USACO08JAN]Telephone Lines S

P1948 [USACO08JAN]Telephone Lines S

概括题意:将图上的 $k$ 条边的边权改为 $0$ 后,求图中所有路径边权最大值的最小值。

考虑二分答案,对于当前的二分值 $x$ ,除了 $k$ 个边权允许使用免费,不允许边权再大于 $x$ ,判断是否存在一条路径从起点到终点。那么,我们需要知道从起点到终点最少使用多少次免费,边权大于 $x$ 的必须使用免费,将边权改为 $1$ ;否则,边权改为 $0$ 。做最短路即可。

查看代码
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#include <iostream>
#include <cstdio>
#include <queue>
#include <climits>
#define INF INT_MAX
using namespace std;
const int N = 1e3 + 10, M = 2e4 + 10;
int n, m, k, dis[N];
int idx = -1, hd[N], nxt[M], edg[M], wt[M], lth[M];
bool vis[N];
void spfa()
{
for (int i = 1; i <= n; i++)
dis[i] = INF;
dis[1] = 0;
queue <int> q;
q.push(1);
vis[1] = true;
while (!q.empty())
{
int t = q.front();
q.pop();
vis[t] = false;
for (int i = hd[t]; ~i; i = nxt[i])
if (dis[t] + wt[i] < dis[edg[i]])
{
dis[edg[i]] = dis[t] + wt[i];
if (!vis[edg[i]])
{
q.push(edg[i]);
vis[edg[i]] = true;
}
}
}
}
bool check(int x)
{
for (int i = 0; i <= idx; i++)
wt[i] = lth[i] > x;
spfa();
return dis[n] <= k;
}
void add(int x, int y, int z)
{
nxt[++idx] = hd[x];
hd[x] = idx;
edg[idx] = y;
lth[idx] = z;
}
int main()
{
int l = 0, r = 0, mid;
cin >> n >> m >> k;
for (int i = 1; i <= n; i++)
hd[i] = -1;
for (int i = 1, a, b, c; i <= m; i++)
{
cin >> a >> b >> c;
r = max(r, c);
add(a, b, c);
add(b, a, c);
}
for (int i = 0; i <= idx; i++)
wt[i] = 1;
spfa();
if (dis[n] == INF)
{
cout << -1;
return 0;
}
while (l < r)
{
mid = l + r >> 1;
if (check(mid))
r = mid;
else
l = mid + 1;
}
cout << l;
return 0;
}