题目链接

分析

英雄方面很简单,跑一遍 Dijkstra 就行了,但是灭火团队就有点麻烦了。

这里可以借助一下最大流的建边来解决这个问题:

我们可以另外找一个点作为起点,然后建立从那个点到每一个团队的起点的边,权值为0,这样就完成了多起点的最短路

恰好我的板子是封装好的 Dijkstra ,我就直接建立两个结构体解决问题,因为点的数量只有 1000 个,空间上已经没有什么顾虑了

AC-Code

#include <bits/stdc++.h>

using namespace std;

#define MAXN 1100
#define MAXM 1000000 #define INF 0x3fffffff //防止后面溢出,这个不能太大 struct Graph {
struct Edge {
long long to, next;
long long cost;
} edge[MAXM];
long long head[MAXN];
long long tot; void init(long long n) {
tot = 0;
memset(head, -1, sizeof(long long) * (n + 1));
} void add_edge(long long from, long long to, long long value) {
edge[tot].to = to;
edge[tot].cost = value;
edge[tot].next = head[from];
head[from] = tot++;
}
}; struct Dijkstra {
long long low_cost[MAXN];
bool vis[MAXN];
long long pre[MAXN]; void solve(long long b, long long e, long long start, Graph &graph) {
for (long long i = b; i < e; i++) {
low_cost[i] = INF;
vis[i] = false;
pre[i] = -1;
}
low_cost[start] = 0;
vis[start] = true;
long long cur_edge = graph.head[start];
while (cur_edge != -1) {
if (!vis[graph.edge[cur_edge].to] &&
low_cost[start] + graph.edge[cur_edge].cost < low_cost[graph.edge[cur_edge].to]) {
low_cost[graph.edge[cur_edge].to] = low_cost[start] + graph.edge[cur_edge].cost;
pre[graph.edge[cur_edge].to] = start;
}
cur_edge = graph.edge[cur_edge].next;
}
for (long long j = b; j < e - 1; j++) {
long long k = -1;
long long Min = INF;
for (long long i = b; i < e; i++) {
if (!vis[i] && low_cost[i] < Min) {
Min = low_cost[i];
k = i;
}
}
if (k == -1)
break;
vis[k] = true;
cur_edge = graph.head[k];
while (cur_edge != -1) {
if (!vis[graph.edge[cur_edge].to] &&
low_cost[k] + graph.edge[cur_edge].cost < low_cost[graph.edge[cur_edge].to]) {
low_cost[graph.edge[cur_edge].to] = low_cost[k] + graph.edge[cur_edge].cost;
pre[graph.edge[cur_edge].to] = k;
}
cur_edge = graph.edge[cur_edge].next;
}
}
}
}; Graph graph;
Dijkstra dijkstra1, dijkstra2;
int k_node[MAXN]; void solve() {
long long t;
cin >> t;
long long v, e, s, k, c;
for (int ts = 0; ts < t; ++ts) {
cin >> v >> e >> s >> k >> c;
graph.init(v + 1);
for (int i = 0; i < k; ++i) {
cin >> k_node[i];
}
long long from, to, value;
for (long long i = 0; i < e; ++i) {
cin >> from >> to >> value;
graph.add_edge(from, to, value);
graph.add_edge(to, from, value);
}
dijkstra1.solve(1, v + 1, s, graph);//第一次跑dijkstra
for (int i = 0; i < k; ++i) {
graph.add_edge(0, k_node[i], 0); // 这里设定超级源点为0,建立从0到每一个团队起点的边,权值为0
}
dijkstra2.solve(0, v + 1, 0, graph);//第二次跑dijkstra long long s_min_max = 0;
for (long long i = 1; i < v + 1; ++i)
s_min_max = max(s_min_max, dijkstra1.low_cost[i]); long long k_min_max = 0;
for (long long i = 1; i < v + 1; ++i)
k_min_max = max(k_min_max, dijkstra2.low_cost[i]); if (s_min_max <= c * k_min_max)//考虑到精度问题,这里用乘法代替
cout << s_min_max << endl;
else
cout << k_min_max << endl;
}
} int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
#ifdef ACM_LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
long long test_index_for_debug = 1;
char acm_local_for_debug;
while (cin >> acm_local_for_debug) {
cin.putback(acm_local_for_debug);
if (test_index_for_debug > 100) {
throw runtime_error("Check the stdin!!!");
}
auto start_clock_for_debug = clock();
cout << "Test " << test_index_for_debug << ":" << endl;
solve();
auto end_clock_for_debug = clock();
cerr << "Test " << test_index_for_debug++ << " Run Time: "
<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
cout << "\n--------------------------------------------------" << endl;
}
#else
solve();
#endif
return 0;
}

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