0x61 图论-最短路

B题 Telephone Lines

https://ac.nowcoder.com/acm/contest/1055/B

中文题面：https://www.luogu.com.cn/problem/P1948

分层图最短路

#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
 
inline int read() {
    int s = 0, w = 1; char ch = getchar();
    while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }
    while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
    return s * w;
}
 
const int N = 2e4 + 7; //节点数
const int M = 2e4 + 7; //路径数
const ll INF = 1e18;
ll d1[N];//d1正向图，d2反向图
struct Node {
    int v;
    ll cost;
    bool operator < (Node b)const {
        return cost > b.cost;
    }
};
vector<Node> e[N];
int n, m, k;
 
void dijkstra(int s, ll d[], int p) {
    priority_queue<Node>  pq;
    fill(d, d + N, INF);
    d[s] = 0;
    pq.push(Node{ s,0 });
    while (!pq.empty()) {
        Node    x = pq.top();
        pq.pop();
        if (d[x.v] < x.cost)    continue; //原先这个节点花费更少 跳过
        for (auto it : e[x.v]) {
            int z = it.cost > p;
            if (d[it.v] > x.cost + z) {
                d[it.v] = x.cost + z;
                pq.push(Node{ it.v,d[it.v] });
            }
        }
    }
}
 
int main() {
    n = read(), m = read(), k = read();
    for (int i = 1; i <= m; ++i) {
        int u = read(), v = read(), w = read();
        e[u].push_back(Node{ v, w });
        e[v].push_back(Node{ u, w });
    }
    int l = 0, r = 1e9, ans = -1;
    while (l <= r) {
        int mid = l + r >> 1;
        dijkstra(1, d1, mid);
        if (d1[n] <= k)  ans = mid, r = mid - 1;
        else   l = mid + 1;
    }
    printf("%d\n", ans);
    return 0;
}

dijkstra+dp

C题 最优贸易

https://ac.nowcoder.com/acm/contest/1055/C

#include<bits/stdc++.h>
using namespace std;
#define sc(n) scanf("%c",&n)
#define sd(n) scanf("%d",&n)
#define pd(n) printf("%d\n", (n))
#define sdd(n,m) scanf("%d %d",&n,&m)
#define sddd(n,m,z) scanf("%d %d %d",&n,&m,&z)
#define pdd(n,m) printf("%d %d\n",n, m)
#define ms(a,b) memset(a,b,sizeof(a))
#define all(c) c.begin(),c.end()
#define pb push_back 
#define fi first 
#define se second 
#define mod(x) ((x)%MOD)
#define lowbit(x) (x & (-x))
#define gcd(a,b) __gcd(a,b)
 
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const int inf = 0x3f3f3f3f;
const int maxn = 100010;
 
struct u {
    int v, len;
};
int n, m, v[maxn], d[maxn * 3 + 1];
vector<u>G[maxn * 3 + 1];
 
void addEdge(int x, int y) {
    G[x].push_back({ y,0 });
    G[x + n].push_back({ y + n,0 });
    G[x + 2 * n].push_back({ y + 2 * n,0 });
    G[x].pb({ y + n,-v[x] });
    G[x + n].pb({ y + 2 * n,v[x] });
}
 
queue<int>q;
bool inq[maxn * 3 + 1];
 
void spfa() {
    ms(d, -inf);
    d[1] = 0; inq[1] = true;
    q.push(1);
    while (q.size()) {
        int tp = q.front(); q.pop();
        inq[tp] = false;
        int len = G[tp].size();
        for (int i = 0; i < len; i++) {
            u x = G[tp][i];
            if (d[x.v] < d[tp] + x.len) {
                d[x.v] = d[tp] + x.len;
                if (inq[x.v] == false) {
                    q.push(x.v);
                    inq[x.v] = true;
                }
            }
        }
    }
}
 
int main() {
    //freopen("in.txt","r",stdin);  //调试用的
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> v[i];
    for (int i = 1, x, y, z; i <= m; i++) {
        cin >> x >> y >> z;
        addEdge(x, y);
        if (z == 2) addEdge(y, x);
    }
    G[n].push_back( { 3 * n + 1, 0 });
    G[n * 3].push_back( { 3 * n + 1, 0 });//超级终点是3*n+1编号
    n = 3 * n + 1; //把n改成超级终点的编号，方便spfa操作
 
    spfa();
    cout << d[n] << endl;
    return 0;
}

D题 Roads and Planes

https://ac.nowcoder.com/acm/contest/1055/d

#include<bits/stdc++.h>
using namespace std;
#define N 25005
#define M 200005
struct Node
{
    int val, pos;
    friend bool operator < (Node x, Node y) {
        return x.val > y.val;
    }
};
struct E { int next, to, dis; } e[M];
int n, m1, m2, s, num, dfn;
int h[N], bel[N], deg[N], dis[N];
bool vis[N];
vector<int> c[N];
queue<int> que;

int read()
{
    int x = 0, f = 1; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }
    while (c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
    return x *= f;
}

void add(int u, int v, int w)
{
    e[++num].next = h[u];
    e[num].to = v;
    e[num].dis = w;
    h[u] = num;
}

void dfs(int x)
{
    bel[x] = dfn, vis[x] = 1, c[dfn].push_back(x);
    for (int i = h[x]; i != 0; i = e[i].next)
        if (!vis[e[i].to]) dfs(e[i].to);
}

void dijkstra(int id)
{
    priority_queue<Node> pue;
    for (int i = 0; i < c[id].size(); i++) pue.push((Node) { dis[c[id][i]], c[id][i] });
    while (pue.size())
    {
        int now = pue.top().pos;
        pue.pop();
        if (vis[now]) continue;
        vis[now] = 1;
        for (int i = h[now]; i != 0; i = e[i].next)
        {
            if (dis[now] + e[i].dis < dis[e[i].to])
            {
                dis[e[i].to] = dis[now] + e[i].dis;
                if (bel[now] == bel[e[i].to]) pue.push((Node) { dis[e[i].to], e[i].to });
            }
            deg[bel[e[i].to]]--;
            if (!deg[bel[e[i].to]] && bel[now] != bel[e[i].to]) que.push(bel[e[i].to]);
        }
    }
}

int main()
{
    cin >> n >> m1 >> m2 >> s;
    for (int i = 1; i <= m1; i++)
    {
        int u = read(), v = read(), w = read();
        add(u, v, w), add(v, u, w);
    }
    for (int i = 1; i <= n; i++)
        if (!vis[i]) ++dfn, dfs(i);
    for (int i = 1; i <= m2; i++)
    {
        int u = read(), v = read(), w = read();
        add(u, v, w);
        deg[bel[v]]++;
    }
    memset(vis, 0, sizeof(vis));
    memset(dis, 0x3f, sizeof(dis));
    dis[s] = 0, que.push(bel[s]);
    for (int i = 1; i <= dfn; i++)
        if (!deg[i]) que.push(i);
    while (que.size())
    {
        int now = que.front();
        que.pop();
        dijkstra(now);
    }
    for (int i = 1; i <= n; i++)
        if (dis[i] >= 1e9) printf("NO PATH\n");
        else printf("%d\n", dis[i]);
}

E题 Sorting It All Out

https://ac.nowcoder.com/acm/contest/1055/e

#include<bits/stdc++.h>
using namespace std;
#define sc(n) scanf("%c",&n)
#define sd(n) scanf("%d",&n)
#define pd(n) printf("%d\n", (n))
#define sdd(n,m) scanf("%d %d",&n,&m)
#define sdcd(n,c,m) scanf("%d%c%d",&n,&c,&m)
#define sddd(n,m,z) scanf("%d %d %d",&n,&m,&z)
#define pdd(n,m) printf("%d %d\n",n, m)
#define ms(a,b) memset(a,b,sizeof(a))
#define mod(x) ((x)%MOD)
#define lowbit(x) (x & (-x))
 
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
const int MOD = 10000007;
const int inf = 0x3f3f3f3f;
const int maxn = 300;
 
int n, m;
char rel[10], ans[maxn];
int dp[maxn][maxn];
 
int judge() {
    for (int k = 1; k <= n; ++k)
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                dp[i][j] |= dp[i][k] & dp[k][j];
 
    int state = 1;
    for (int i = 1; i <= n; i++) {
        if (dp[i][i]) return -1;
        for (int j = i + 1; j <= n; j++) {
            if (!(dp[i][j] | dp[j][i])) state = 0;
        }
    }
    return state;
}
 
void get_ans()
{
    for (int i = 1; i <= n; i++) {
        int pos = n;
        for (int j = 1; j <= n; j++)
            if (dp[i][j]) pos--;
        ans[pos] = char(i + 'A' - 1);
    }
    ans[n + 1] = 0;
}
 
int main() {
    //freopen("in.txt", "r", stdin);
    //ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    while (cin >> n >> m && n && m) {
        ms(dp, 0);
        int state = 0, success_ans = 0, failure_ans = 0;
        for (int i = 1; i <= m; i++) {
            scanf("%s", rel);
            if (state) continue;
            int x = rel[0] - 'A' + 1;
            int y = rel[2] - 'A' + 1;
            dp[x][y] = 1;
            state = judge();
            if (state == 1 && success_ans == 0) success_ans = i;
            if (state == -1) failure_ans = i;
        }
 
        if (state == 1) {
            get_ans();
            printf("Sorted sequence determined after %d relations: %s.\n", success_ans, ans + 1);
        }
        else if (state == -1) {
            printf("Inconsistency found after %d relations.\n", failure_ans);
        }
        else {
            printf("Sorted sequence cannot be determined.\n");
        }
    }
    return 0;
}

F题 Sightseeing trip

https://ac.nowcoder.com/acm/contest/1055/f

#include<bits/stdc++.h>
using namespace std;
#define ms(a,b) memset(a,b,sizeof a)
const int maxn = 310;
const int inf = 0x3f3f3f3f;
int a[maxn][maxn], d[maxn][maxn], pos[maxn][maxn];
int n, m, ans = inf;
vector<int> path;
 
void get_path(int x, int y) {
    if (pos[x][y] == 0)return;
    get_path(x, pos[x][y]);
    path.push_back(pos[x][y]);
    get_path(pos[x][y], y);
}
 
int main() {
    //freopen("in.txt", "r", stdin);
    ios::sync_with_stdio(false), cin.tie(0);
    int x, y, z;
    cin >> n >> m;
    ms(a, 0x3f);
    for (int i = 1; i <= n; ++i)a[i][i] = 0;
    while (m--) {
        cin >> x >> y >> z;
        a[x][y] = a[y][x] = min(a[y][x], z);
    }
 
    memcpy(d, a, sizeof a);
    for (int k = 1; k <= n; k++) {
        for (int i = 1; i < k; i++)
            for (int j = i + 1; j < k; j++)
                if ((long long)d[i][j] + a[j][k] + a[k][i] < ans) {
                    ans = d[i][j] + a[j][k] + a[k][i];
                    path.clear();
                    path.push_back(i);
                    get_path(i, j);
                    path.push_back(j);
                    path.push_back(k);
                }
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                if (d[i][j] > d[i][k] + d[k][j]) {
                    d[i][j] = d[i][k] + d[k][j];
                    pos[i][j] = k;
                }
    }
    if (ans == 0x3f3f3f3f) {
        puts("No solution.");
        return 0;
    }
    for (auto p : path)
        cout << p << " ";
    cout << endl;
}

G题 Cow Relays

https://ac.nowcoder.com/acm/contest/1055/g

思路：

给出一张无向连通图，求S到E经过k条边的最短路。

矩阵我不熟,看了大佬的一个式子:

把经过xx个点的最短路的邻接矩阵\(X\)和经过\(y\)个点的最短路的邻接矩阵\(Y\)合并的式子为:

\[Ai,j=min(Ai,j,Xi,k+Yk,j) \]

把输入转成邻接矩阵后,这个邻接矩阵可以看作恰好经过一个点的最短路,然后转移n−1n−1次就可以了

矩阵相乘时,需要使用快速幂优化

AC代码

#include<bits/stdc++.h>
using namespace std;
int num[1000005], n, s, t, e, tol, x, y, z;
struct map {
    int data[500][500];
    map operator*(const map& other) const {
        map c;
        for (int k = 1; k <= tol; k++)
            for (int i = 1; i <= tol; i++)
                for (int j = 1; j <= tol; j++)
                    c.data[i][j] =
                    min(c.data[i][j], data[i][k] + other.data[k][j]);
        return c;
    }
    map() { memset(data, 0x3f3f3f3f, sizeof(data)); }
} dis, ans;
inline int input() { int t; scanf("%d", &t);  return t; }
int main() {
    n = input() - 1, t = input(), s = input(), e = input();
    for (int i = 1; i <= t; i++) {
        x = input();
        if (!num[y = input()])num[y] = ++tol;
        if (!num[z = input()])num[z] = ++tol;
        dis.data[num[y]][num[z]] = dis.data[num[z]][num[y]] = x;
    }
    ans = dis;
    while (n) (n & 1) && (ans = ans * dis, 0), dis = dis * dis, n >>= 1;
    printf("%d", ans.data[num[s]][num[e]]);
}