1 介绍

本专题用来介绍使用最短路算法（spfa或dijkstra）与其它算法（dfs、二分、拓扑排序、动态规划等等）结合起来解题。

2 训练

题目1：1135新年好

C++代码如下，

//版本1，使用vector来存储图，会有超时风险
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>using namespace std;const int N = 50010;
int n, m;
int dist[6][N];
bool st[N];
vector<vector<pair<int, int>>> g; //使用vector可能有超时分险
int source[6];void spfa(int start, int dist[]) {memset(dist, 0x3f, N * 4);memset(st, 0, sizeof st);dist[start] = 0;queue<int> q;q.push(start);st[start] = true;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 dfs(int u, int start, int distance) {if (u > 5) return distance;int res = 0x3f3f3f3f;for (int i = 1; i <= 5; ++i) {if (!st[i]) {int b = source[i];st[i] = true;res = min(res, dfs(u + 1, i, distance + dist[start][b]));st[i] = false; }}return res;
}int main() {cin >> n >> m;g.resize(n + 10);source[0] = 1;for (int i = 1; i <= 5; ++i) cin >> source[i];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);}for (int i = 0; i < 6; ++i) spfa(source[i], dist[i]);memset(st, 0, sizeof st);int res = dfs(1, 0, 0);cout << res << endl;return 0;   
}

//版本2，使用h,e,ne,w数组来存储图
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>using namespace std;typedef pair<int, int> PII;const int N = 50010, M = 200010, INF = 0x3f3f3f3f;int n, m;
int h[N], e[M], w[M], ne[M], idx;
int q[N], dist[6][N];
int source[6];
bool st[N];void add(int a, int b, int c) {e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}void dijkstra(int start, int dist[]) {memset(dist, 0x3f, N * 4);dist[start] = 0;memset(st, 0, sizeof st);priority_queue<PII, vector<PII>, greater<PII>> heap;heap.push({0, start});while (heap.size()) {auto t = heap.top();heap.pop();int ver = t.second;if (st[ver]) continue;st[ver] = true;for (int i = h[ver]; ~i; i = ne[i]) {int j = e[i];if (dist[j] > dist[ver] + w[i]) {dist[j] = dist[ver] + w[i];heap.push({dist[j], j});}}}
}int dfs(int u, int start, int distance) {if (u > 5) return distance;int res = INF;for (int i = 1; i <= 5; ++i) {if (!st[i]) {int next = source[i];st[i] = true;res = min(res, dfs(u + 1, i, distance + dist[start][next]));st[i] = false; }}return res;
}int main() {scanf("%d%d", &n, &m);source[0] = 1;for (int i = 1; i <= 5; ++i) scanf("%d", &source[i]);memset(h, -1, sizeof h);while (m--) {int a, b, c;scanf("%d%d%d", &a, &b, &c);add(a, b, c), add(b, a, c);}for (int i = 0; i < 6; ++i) dijkstra(source[i], dist[i]);memset(st, 0, sizeof st);printf("%d\n", dfs(1, 0, 0));return 0;
}

题目2：340通信线路

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>
#include <deque>using namespace std;const int N = 1010, M = 20010;
int n, m, k;
int h[N], e[M], w[M], ne[M], idx;
int dist[N];
deque<int> q;
bool st[N];void add(int a, int b, int c) {e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}bool check(int bound) {memset(dist, 0x3f, sizeof dist);memset(st, 0, sizeof st);q.push_back(1);dist[1] = 0;while (q.size()) {int t = q.front();q.pop_front();if (st[t]) continue;st[t] = true;for (int i = h[t]; ~i; i = ne[i]) {int j = e[i], x = w[i] > bound;if (dist[j] > dist[t] + x) {dist[j] = dist[t] + x;if (!x) q.push_front(j);else q.push_back(j);}}}return dist[n] <= k;
}int main() {cin >> n >> m >> k;memset(h, -1, sizeof h);while (m--) {int a, b, c;cin >> a >> b >> c;add(a, b, c), add(b, a, c);}int l = 0, r = 1e6 + 1;while (l < r) {int mid = l + r >> 1;if (check(mid)) r = mid;else l = mid + 1;}if (r == 1e6 + 1) cout << -1 << endl;else cout << r << endl;return 0;
}

题目3：342道路与航线

C++代码如下，

#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>#define x first
#define y secondusing namespace std;typedef pair<int, int> PII;const int N = 25010, M = 150010, INF = 0x3f3f3f3f;
int n, mr, mp, S;
int id[N];
int h[N], e[M], w[M], ne[M], idx;
int dist[N], din[N];
vector<int> block[N];
int bcnt;
bool st[N];
queue<int> q;void add(int a, int b, int c) {e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}void dfs(int u, int bid) {id[u] = bid, block[bid].push_back(u);for (int i = h[u]; ~i; i = ne[i]) {int j = e[i];if (!id[j]) {dfs(j, bid);}}
}void dijkstra(int bid) {priority_queue<PII, vector<PII>, greater<PII>> heap;for (auto u : block[bid]) {heap.push({dist[u], u});}while (heap.size()) {auto t = heap.top();heap.pop();int ver = t.y, distance = t.x;if (st[ver]) continue;st[ver] = true;for (int i = h[ver]; ~i; i = ne[i]) {int j = e[i];if (id[j] != id[ver] && --din[id[j]] == 0) q.push(id[j]);if (dist[j] > dist[ver] + w[i]) {dist[j] = dist[ver] + w[i];if (id[j] == id[ver]) heap.push({dist[j], j});}}}
}void topsort() {memset(dist, 0x3f, sizeof dist);dist[S] = 0;for (int i = 1; i <= bcnt; ++i) {if (!din[i]) {q.push(i);}}while (q.size()) {int t = q.front();q.pop();dijkstra(t);}}int main() {cin >> n >> mr >> mp >> S;memset(h, -1, sizeof h);while (mr--) {int a, b, c;cin >> a >> b >> c;add(a, b, c), add(b, a, c);}for (int i = 1; i <= n; ++i) {if (!id[i]) {bcnt++;dfs(i, bcnt);}}while (mp--) {int a, b, c;cin >> a >> b >> c;din[id[b]]++;add(a, b, c);}topsort();for (int i = 1; i <= n; ++i) {if (dist[i] > INF / 2) cout << "NO PATH" << endl;else cout << dist[i] << endl;}return 0;
}

题目4：341最优贸易

C++代码如下，

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;const int N = 100010, M = 2000010;
int n, m;
int w[N];
int hs[N], ht[N], e[M], ne[M], idx;
int dmin[N], dmax[N];
int q[N];
bool st[N];void add(int h[], int a, int b) {e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}void spfa(int h[], int dist[], int type) {int hh = 0, tt = 1;if (type == 0) {memset(dist, 0x3f, sizeof dmin);dist[1] = w[1];q[0] = 1;} else {memset(dist, -0x3f, sizeof dmax);dist[n] = w[n];q[0] = n;}while (hh != tt) {int t = q[hh++];if (hh == N) hh = 0;st[t] = false;for (int i = h[t]; ~i; i = ne[i]) {int j = e[i];if (type == 0 && dist[j] > min(dist[t], w[j]) || type == 1 && dist[j] < max(dist[t], w[j])) {if (type == 0) dist[j] = min(dist[t], w[j]);else dist[j] = max(dist[t], w[j]);if (!st[j]) {q[tt++] = j;if (tt == N) tt = 0;st[j] = true;}}}}
}int main() {scanf("%d%d", &n, &m);for (int i = 1; i <= n; ++i) scanf("%d", &w[i]);memset(hs, -1, sizeof hs);memset(ht, -1, sizeof ht);while (m--) {int a, b, c;scanf("%d%d%d", &a, &b, &c);add(hs, a, b), add(ht, b, a);if (c == 2) add(hs, b, a), add(ht, a, b);}spfa(hs, dmin, 0);spfa(ht, dmax, 1);int res = 0;for (int i = 1; i <= n; ++i) res = max(res, dmax[i] - dmin[i]);printf("%d\n", res);return 0;
}