1 介绍
本专题用来记录使用。。。。
2 训练
题目1：1137选择最佳线路
C++代码如下，
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 1010, M = 20010;
int n, m;
int dist[N];
int st[N];
vector<vector<pair<int,int>>> g;
vector<int> snodes;
int enode;
int w;
void spfa() {
//cout << " ====== " << endl;
memset(dist, 0x3f, sizeof dist);
memset(st, 0, sizeof st);
queue<int> q;
for (auto snode : snodes) {
//cout << "snode = " << snode << endl;
dist[snode] = 0;
q.push(snode);
st[snode] = true; //已经在队列中了
}
while (!q.empty()) {
auto t = q.front();
q.pop();
st[t] = false;
for (auto [b, w] : g[t]) {
if (dist[b] > dist[t] + w) {
dist[b] = dist[t] + w;
if (!st[b]) {
q.push(b);
st[b] = true;
}
}
}
}
return;
}
int main() {
while (cin >> n >> m >> enode) {
g.clear();
g.resize(n + 10);
for (int i = 0; i < m; ++i) {
int a, b, c;
cin >> a >> b >> c;
g[a].emplace_back(b, c);
}
cin >> w;
snodes.resize(w);
for (int i = 0; i < w; ++i) cin >> snodes[i];
spfa();
if (dist[enode] == 0x3f3f3f3f) cout << "-1" << endl;
else cout << dist[enode] << endl;
}
return 0;
}
题目2：1131拯救大兵瑞恩
C++代码如下，
#include <cstring>
#include <iostream>
#include <algorithm>
#include <deque>
#include <set>
#define x first
#define y second
using namespace std;
typedef pair<int,int> PII;
const int N = 11, M = 360, P = 1 << 10;
int n, m, k, p;
int h[N * N], e[M], w[M], ne[M], idx;
int g[N][N], key[N * N];
int dist[N * N][P];
bool st[N * N][P];
set<PII> edges;
void add(int a, int b, int c) {
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}
void build() {
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
for (int u = 0; u < 4; ++u) {
int x = i + dx[u], y = j + dy[u];
if (!x || x > n || !y || y > m) continue;
int a = g[i][j], b = g[x][y];
if (!edges.count({a,b})) add(a, b, 0);
}
}
}
return;
}
int bfs() {
memset(dist, 0x3f, sizeof dist);
dist[1][0] = 0;
deque<PII> q;
q.push_back({1, 0});
while (q.size()) {
PII t = q.front();
q.pop_front();
if (st[t.x][t.y]) continue;
st[t.x][t.y] = true;
if (t.x == n * m) return dist[t.x][t.y];
if (key[t.x]) {
int state = t.y | key[t.x];
if (dist[t.x][state] > dist[t.x][t.y]) {
dist[t.x][state] = dist[t.x][t.y];
q.push_front({t.x, state});
}
}
for (int i = h[t.x]; ~i; i = ne[i]) {
int j = e[i];
if (w[i] && !(t.y >> w[i] - 1 & 1)) continue;
if (dist[j][t.y] > dist[t.x][t.y] + 1) {
dist[j][t.y] = dist[t.x][t.y] + 1;
q.push_back({j, t.y});
}
}
}
return -1;
}
int main() {
cin >> n >> m >> p >> k;
for (int i = 1, t = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
g[i][j] = t++;
}
}
memset(h, -1, sizeof h);
while (k--) {
int x1, y1, x2, y2, c;
cin >> x1 >> y1 >> x2 >> y2 >> c;
int a = g[x1][y1], b = g[x2][y2];
edges.insert({a, b}), edges.insert({b, a});
if (c) add(a, b, c), add(b, a, c);
}
build();
int s;
cin >> s;
while (s--) {
int x, y, c;
cin >> x >> y >> c;
key[g[x][y]] |= 1 << c - 1;
}
cout << bfs() << endl;
return 0;
}
题目3：1134最短路计数
C++代码如下，
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int N = 1e5 + 10, mod = 100003;
int n, m;
vector<vector<int>> g;
int dist[N];
int cnt[N];
void bfs() {
memset(dist, 0x3f, sizeof dist);
queue<int> q;
q.push(1);
cnt[1] = 1;
dist[1] = 0;
while (!q.empty()) {
auto t = q.front();
q.pop();
for (auto b : g[t]) {
if (dist[b] > dist[t] + 1) {
dist[b] = dist[t] + 1;
cnt[b] = cnt[t];
q.push(b);
} else if (dist[b] == dist[t] + 1) {
cnt[b] = (cnt[b] + cnt[t]) % mod;
}
}
}
return;
}
int main() {
cin >> n >> m;
g.resize(n + 10);
for (int i = 0; i < m; ++i) {
int a, b;
cin >> a >> b;
g[a].emplace_back(b);
g[b].emplace_back(a);
}
bfs();
for (int i = 1; i <= n; ++i) {
cout << cnt[i] << endl;
}
return 0;
}
题目4：383观光
C++代码如下，
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <tuple>
#include <vector>
using namespace std;
typedef tuple<int,int,int> TIII; //distance,node,type
const int N = 1010;
int dist[N][2];
int cnt[N][2];
bool st[N][2];
int n, m;
int S, F;
vector<vector<pair<int,int>>> g;
void dijkstra() {
memset(dist, 0x3f, sizeof dist);
memset(cnt, 0, sizeof cnt);
memset(st, 0, sizeof st);
priority_queue<TIII, vector<TIII>, greater<TIII>> hp;
dist[S][0] = 0, cnt[S][0] = 1;
hp.push({0, S, 0});
while (!hp.empty()) {
int distance, node, type;
tie(distance, node, type) = hp.top();
hp.pop();
int count = cnt[node][type];
if (st[node][type]) continue;
st[node][type] = true;
for (auto [b, w] : g[node]) {
if (dist[b][0] > distance + w) {
dist[b][1] = dist[b][0], cnt[b][1] = cnt[b][0];
hp.push(make_tuple(dist[b][1], b, 1));
dist[b][0] = distance + w, cnt[b][0] = count;
hp.push(make_tuple(dist[b][0], b, 0));
} else if (dist[b][0] == distance + w) cnt[b][0] += count;
else if (dist[b][1] > distance + w) {
dist[b][1] = distance + w, cnt[b][1] = count;
hp.push(make_tuple(dist[b][1], b, 1));
} else if (dist[b][1] == distance + w) cnt[b][1] += count;
}
}
return;
}
int main() {
int cases;
cin >> cases;
while (cases--) {
cin >> n >> m;
g.clear();
g.resize(n + 10);
for (int i = 0; i < m; ++i) {
int a, b, c;
cin >> a >> b >> c;
g[a].emplace_back(b, c);
}
cin >> S >> F;
dijkstra();
int res = cnt[F][0];
if (dist[F][0] + 1 == dist[F][1]) res += cnt[F][1];
cout << res << endl;
}
return 0;
}