题意:收过路费.如果最后的收费小于3或不能达到,输出'?'.否则输出到n点最小的过路费
分析:关键权值可为负,如果碰到负环是,小于3的约束条件不够,那么在得知有负环时,把这个环的点都标记下,DFS实现.
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std; const int N = 2e2 + 5;
const int E = 1e4 + 5;
const int INF = 0x3f3f3f3f;
struct Edge {
int v, w, nex;
Edge() {}
Edge(int v, int w, int nex) : v (v), w (w), nex (nex) {}
bool operator < (const Edge &r) const {
return w > r.w;
}
}edge[E];
int head[N];
int d[N];
int cnt[N];
int a[N];
bool vis[N], vis2[N];
int n, m, e; void init() {
memset (head, -1, sizeof (head));
e = 0;
} void add_edge(int u, int v, int w) {
edge[e] = Edge (v, w, head[u]);
head[u] = e++;
} void DFS(int u) {
vis2[u] = true;
for (int i=head[u]; ~i; i=edge[i].nex) {
int v = edge[i].v;
if (!vis2[v]) {
DFS (v);
}
}
} void SPFA(int s) {
memset (cnt, 0, sizeof (cnt));
memset (vis, false, sizeof (vis));
memset (vis2, false, sizeof (vis2));
memset (d, INF, sizeof (d));
d[s] = 0; cnt[s] = 0; vis[s] = true;
queue<int> que; que.push (s);
while (!que.empty ()) {
int u = que.front (); que.pop ();
vis[u] = false;
for (int i=head[u]; ~i; i=edge[i].nex) {
int v = edge[i].v, w = edge[i].w;
if (vis2[v]) continue;
if (d[v] > d[u] + w) {
d[v] = d[u] + w;
if (!vis[v]) {
vis[v] = true; que.push (v);
if (++cnt[v] > n) {
DFS (v);
}
}
}
}
}
} int cal(int i, int j) {
int ret = a[i] - a[j];
ret = ret * ret * ret;
return ret;
} int main(void) {
int T, cas = 0; scanf ("%d", &T);
while (T--) {
init ();
scanf ("%d", &n);
for (int i=1; i<=n; ++i) {
scanf ("%d", &a[i]);
}
scanf ("%d", &m);
for (int u, v, i=1; i<=m; ++i) {
scanf ("%d%d", &u, &v);
add_edge (u, v, cal (v, u));
}
SPFA (1);
int q; scanf ("%d", &q);
printf ("Case %d:\n", ++cas);
while (q--) {
int x; scanf ("%d", &x);
if (d[x] == INF || d[x] < 3 || vis2[x]) puts ("?");
else printf ("%d\n", d[x]);
}
} return 0;
}