POJ 3114 Countries in War（强连通+最短路）

POJ 3114 Countries in War

题目链接

题意：给定一个有向图，强连通分支内传送不需要花费，其他有一定花费，每次询问两点的最小花费

思路：强连通缩点后求最短路即可

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;

const int MAXNODE = 505;
const int MAXEDGE = 255005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
int u, v;
Type dist;
Edge() {}
Edge(int u, int v, Type dist) {
this->u = u;
this->v = v;
this->dist = dist;
}
};

struct HeapNode {
Type d;
int u;
HeapNode() {}
HeapNode(Type d, int u) {
this->d = d;
this->u = u;
}
bool operator < (const HeapNode& c) const {
return d > c.d;
}
};

struct Dijkstra {
int n, m;
Edge edges[MAXEDGE];
int first[MAXNODE];
int next[MAXEDGE];
bool done[MAXNODE];
Type d[MAXNODE];

void init(int n) {
this->n = n;
memset(first, -1, sizeof(first));
m = 0;
}

void add_Edge(int u, int v, Type dist) {
edges[m] = Edge(u, v, dist);
next[m] = first[u];
first[u] = m++;
}

int dijkstra(int s, int t) {
priority_queue<HeapNode> Q;
for (int i = 0; i < n; i++) d[i] = INF;
d[s] = 0;
memset(done, false, sizeof(done));
Q.push(HeapNode(0, s));
while (!Q.empty()) {
HeapNode x = Q.top(); Q.pop();
int u = x.u;
if (u == t) return d[t];
if (done[u]) continue;
done[u] = true;
for (int i = first[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (d[e.v] > d[u] + e.dist) {
d[e.v] = d[u] + e.dist;
Q.push(HeapNode(d[e.v], e.v));
}
}
}
return -1;
}
} gao;

const int N = 505;

int n, m;
vector<Edge> g
;

int pre
, dfn
, dfs_clock, sccn, sccno
;
stack<int> S;

void dfs_scc(int u) {
pre[u] = dfn[u] = ++dfs_clock;
S.push(u);
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i].v;
if (!pre[v]) {
dfs_scc(v);
dfn[u] = min(dfn[u], dfn[v]);
} else if (!sccno[v]) dfn[u] = min(dfn[u], pre[v]);
}
if (pre[u] == dfn[u]) {
sccn++;
while (1) {
int x = S.top(); S.pop();
sccno[x] = sccn;
if (x == u) break;
}
}
}

void find_scc() {
dfs_clock = sccn = 0;
memset(sccno, 0, sizeof(sccno));
memset(pre, 0, sizeof(pre));
for (int i = 1; i <= n; i++)
if (!pre[i]) dfs_scc(i);
}

int main() {
while (~scanf("%d%d", &n, &m) && n) {
for (int i = 1; i <= n; i++) g[i].clear();
int u, v, w;
while (m--) {
scanf("%d%d%d", &u, &v, &w);
g[u].push_back(Edge(u, v, w));
}
find_scc();
gao.init(n);
for (int u = 1; u <= n; u++) {
for (int j = 0; j < g[u].size(); j++) {
int v = g[u][j].v;
if (sccno[u] == sccno[v]) continue;
gao.add_Edge(sccno[u] - 1, sccno[v] - 1, g[u][j].dist);
}
}
int q;
scanf("%d", &q);
while (q--) {
scanf("%d%d", &u, &v);
int tmp = gao.dijkstra(sccno[u] - 1, sccno[v] - 1);
if (tmp == -1) printf("Nao e possivel entregar a carta\n");
else printf("%d\n", tmp);
}
printf("\n");
}
return 0;
}