1 介绍

本博客用来记录使用dijkstra算法或spfa算法求解最短路问题的题目。

2 训练

题目1：1129热浪

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>using namespace std;const int N = 2510;
int n, m;
vector<vector<pair<int,int>>> g; //first表示next_node，second表示w
int dist[N];
bool st[N];
int snode, enode;void dijkstra() {memset(dist, 0x3f, sizeof dist);dist[snode] = 0;priority_queue<pair<int, int>, vector<pair<int,int>>, greater<pair<int,int>>> h;h.push(make_pair(0, snode));while (!h.empty()) {//确定当前结点中，不在集合s且距离结点snode最近的结点。记作cnodeauto t = h.top();h.pop();int cdist = t.first, cnode = t.second;if (st[cnode]) continue; //如果cnode已经被确定是最小路径上的结点了，则跳过st[cnode] = true; //将它加入到集合中for (auto [next_node, w] : g[cnode]) {if (dist[next_node] > cdist + w) {dist[next_node] = cdist + w;h.push(make_pair(dist[next_node], next_node));}}}return;
}int main() {cin >> n >> m >> snode >> enode;g.resize(n + 10);for (int i = 1; i <= m; ++i) {int a, b, c;cin >> a >> b >> c;g[a].emplace_back(b, c);g[b].emplace_back(a, c);}//求snode到enode的最短距离dijkstra();cout << dist[enode] << endl;return 0;
}

题目2：1128信使

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>using namespace std;const int N = 110;
int n, m;
int d[N];
bool st[N];
vector<vector<pair<int,int>>> g;void dijkstra() {memset(d, 0x3f, sizeof d);d[1] = 0;priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> hp; //小根堆hp.push(make_pair(0, 1)); //first表示距离，second表示结点while (!hp.empty()) {auto t = hp.top(); //找到未在集合中，距离最小的结点hp.pop();int a = t.second;if (st[a]) continue; //已经用d[a]更新过了，将它放入集合中st[a] = true;for (auto [b, w] : g[a]) {if (d[b] > d[a] +w) { //d[b]此时比较大，用一个更小值来更新它。d[b] = d[a] + w;hp.push(make_pair(d[b], b));}}}return;
}int main() {cin >> n >> m;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);g[b].emplace_back(a, c);}dijkstra();int res = 0; //求最大值for (int i = 1; i <= n; ++i) res = max(res, d[i]);if (res == 0x3f3f3f3f) {res = -1;}cout << res << endl;return 0;
}

题目3：1127香甜的黄油

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>using namespace std;const int N = 810;
int cow, n, m;
int cnt[N];
int d[N];
bool st[N];
vector<vector<pair<int,int>>> g;void spfa(int start) {//起点为startmemset(d, 0x3f, sizeof d);memset(st, 0, sizeof st);d[start] = 0;queue<int> q;q.push(start);st[start] = true; //结点start在队列中while (!q.empty()) {int t = q.front();q.pop();st[t] = false; //结点t不在队列中了for (auto [b, w] : g[t]) {if (d[b] > d[t] + w) {d[b] = d[t] + w;if (!st[b]) {q.push(b);st[b] = true;}}}}return;
}int main() {cin >> cow >> n >> m;g.resize(n + 10);for (int i = 1; i <= cow; ++i) {int a;cin >> a; //每头牛所在的牧场cnt[a]++;}for (int i = 1; i <= m; ++i) {int a, b, c;cin >> a >> b >> c;g[a].emplace_back(b, c);g[b].emplace_back(a, c);}//spfa()算法 //o(m)时间复杂度，不会被超时long long res = INT_MAX;for (int i = 1; i <= n; ++i) {//第i个牧场作为放糖点spfa(i);long long t = 0; for (int j = 1; j <= n; ++j) t += cnt[j] * d[j];res = min(res, t);}cout << res << endl;return 0;
}

题目4：1126最小花费

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>using namespace std;const int N = 2010;
int n, m;
int snode, enode;
double d[N]; //求最大距离，最大利率
bool st[N]; //是否使用它来更新过
vector<vector<pair<int,int>>> g;void dijkstra() {//d的初始化for (int i = 1; i <= n; ++i) d[i] = 0.0; //初始成0.0memset(st, 0, sizeof st);d[snode] = 1.0;priority_queue<pair<double,int>> hp; //大根堆hp.push(make_pair(1.0, snode));while (!hp.empty()) {auto t = hp.top();hp.pop();int a = t.second;if (st[a]) continue;st[a] = true;for (auto [b, w] : g[a]) {if (d[b] < d[a] * 0.01 * (100 - w)) {d[b] = d[a] * 0.01 * (100 - w);hp.push(make_pair(d[b], b));}}}return;
}int main() {cin >> n >> m;g.resize(n + 10);for (int i = 1; i <= m; ++i) {int a, b, c;cin >> a >> b >> c;g[a].emplace_back(b, c);g[b].emplace_back(a, c);}cin >> snode >> enode;dijkstra();double res = 100.0 / d[enode];printf("%.8f\n", res);return 0;
}

题目5：920最优乘车

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>
#include <sstream>
#include <vector>
#include <queue>using namespace std;const int N = 510;
int n, m;
vector<vector<int>> g;
bool st[N];
int dist[N];void bfs() {memset(dist, 0x3f, sizeof dist);queue<int> q;q.push(1);dist[1] = 0;st[1] = true;while (!q.empty()) {auto t = q.front();q.pop();//t可以走到哪儿for (auto b : g[t]) {if (!st[b]) {dist[b] = dist[t] + 1;q.push(b);st[b] = true;}}}return;
}int main() {cin >> m >> n;g.resize(n + 10);string line;getline(cin, line);for (int i = 0; i < m; ++i) {getline(cin, line);stringstream ssin(line);vector<int> nodes;int node = -1;while (ssin >> node) {nodes.emplace_back(node);}for (int i = 0; i < nodes.size(); ++i) {for (int j = i + 1; j < nodes.size(); ++j) {g[nodes[i]].emplace_back(nodes[j]);}}}bfs();if (dist[n] == 0x3f3f3f3f) puts("NO");else cout << max(dist[n] - 1, 0) << endl;return 0;
}

题目6：903昂贵的聘礼

C++代码如下，

#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;const int N = 110, INF = 0x3f3f3f3f;
int n, m;
int w[N][N], level[N];
int dist[N];
bool st[N];int dijkstra(int down, int up) {memset(dist, 0x3f, sizeof dist);memset(st, 0, sizeof st);dist[0] = 0;for (int i = 1; i <= n + 1; ++i) {int t = -1;for (int j = 0; j <= n; ++j) {if (!st[j] && (t == -1 || dist[t] > dist[j])) {t = j;}}st[t] = true;for (int j = 1; j <= n; ++j) {if (level[j] >= down && level[j] <= up) {dist[j] = min(dist[j], dist[t] + w[t][j]);}}}return dist[1];
}int main() {cin >> m >> n;memset(w, 0x3f, sizeof w);for (int i = 1; i <= n; ++i) w[i][i] = 0;for (int i = 1; i <= n; ++i) {int price, cnt;cin >> price >> level[i] >> cnt;w[0][i] = min(price, w[0][i]);while (cnt--) {int id, cost;cin >> id >> cost;w[id][i] = min(w[id][i], cost);}}int res = INF;for (int i = level[1] - m; i <= level[1]; ++i) res = min(res, dijkstra(i, i + m));cout << res << endl;return 0;
}