3.1.2 单源最短路的综合应用

这章主要讲解单源最短路和其他算法结合
和dfs, 二分, dp, 拓扑排序的结合
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AcWing 1135. 新年好

分析

最短路和dfs的结合

code

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int N = 500010, M = 200010, INF = 0x3f3f3f3f;
typedef pair<int, int> PII;
int h[N], e[M], w[M], ne[M], idx;
int source[6], dist[6][N];
int n, m;
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 s, int dist[]){
    
    
   memset(st, 0, sizeof st);
   memset(dist, 0x3f, N * 4);
   priority_queue<PII, vector<PII>, greater<PII>> heap;
   dist[s] = 0;
   heap.push({
    
    0, s});
   
   while (heap.size()){
    
    
       auto t = heap.top(); heap.pop();
       int ver = t.second, distance = t.first;
       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]){
    
     // 如果没有遍历过
           int next = source[i];
           st[i] = true;
           res = min(res, dfs(u + 1, i, distance + dist[start][next]));
           st[i] = false; // 恢复现场
       }

   return res;
}

int main(){
    
    
   cin >> n >> m;
   source[0] = 1; // 起点1 一定要初始化
   
   for (int i = 1; i <= 5; i ++ ) {
    
    
       int x;
       cin >> x;
       source[i] = x;
   }
   memset(h, -1, sizeof h);
   
   for (int i = 1; i <= m; i ++ ){
    
    
       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)); // 1.已经遍历过几个点, 从start 开始 
   
   return 0;
}

AcWing 340. 通信线路

分析

根据题意, 可以发现我们在图中找第k大的边, 不过下面的分析思路有点太神奇了
二分求出: 从1走到N, 最少经过长度大于x的边的数量是否<= k,
可以发现如果x是答案, 那么经过x右边长度的边的数量一定满足,
经过x左边数量的边的数量, 一定不满足,
因为如果存在x1 < x且 经过大于x1的边的数量也<= k, 那么与x是答案,矛盾.
因此可以用二分
有了二分以后, 可以将所有边分类, 如果边的长度大于x, 则权值为1, 否则边权为0.
权值为0, 1的最短路问题可以用双端队列bfs求解
注意: 二分边界可以取到0, 砍掉所有边, 取到1e6 + 1, 可以用来区分图是否连通, 因为题目最大值x = 1e6, 如果最后二分返回1e6 + 1, 说明无解.
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灰之魔女题解
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code

#include <iostream>
#include <cstring>
#include <deque>
using namespace std;
const int N = 1010, M = 10010 * 2;
int n, m, k;
int h[N], e[M], w[M], ne[M], idx;
int st[N];
int dist[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(st, 0, sizeof st);
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    deque<int> q;
    q.push_back(1);
    
    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], v = w[i] > bound;
            if (dist[j] > dist[t] + v){
    
    
                dist[j] = dist[t] + v;
                if (!v) q.push_front(j);
                else q.push_back(j);
            }
        }
    }
    return dist[n] <= k;
    
}

int main(){
    
    
    memset(h, -1, sizeof h);
    cin >> n >> m >> k;
    
    for (int i = 1; i <= m; i ++ ){
    
    
        int a, b, c;
        scanf("%d%d%d", &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) r = -1;
    cout << r << endl;
    
    return 0;
}

AcWing 342. 道路与航线

分析

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code

#include <iostream>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
const int N = 25010, M = 50010 * 3, INF = 0x3f3f3f3f; // 道路双向 * 2 + 航线, 因此是3倍

#define x first
#define y second
typedef pair<int, int> PII;
int h[N], e[M], w[M], ne[M], idx;
int n, mr, mp, S;
int dist[N], din[N];
int id[N], bcnt; // id[i] 返回i点所在的连通块id
vector<int> block[N];
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]); // q是全局变量, 表示团子队列
            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;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c), add(b, a, c);
    }
    // flood-fill 找连通块
    for (int i = 1; i <= n; i ++ )
        if (!id[i]){
    
    
            bcnt ++;
            dfs(i, bcnt);
        }
        
    while (mp --){
    
    
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        din[id[b]] ++;
        add(a, b, c);
    }
    
    topsort();
    
    for (int i = 1; i <= n; i ++ )
        if (dist[i] > INF / 2) puts("NO PATH");
        else cout << dist[i] << endl;
        
    return 0;

}

AcWing 341. 最优贸易

分析

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code

#include <iostream>
#include <cstring>
using namespace std;
const int N = 1e5 + 10, M = 500010 * 2;
int hs[N], ht[N], e[M], w[M], ne[M], idx;
int n, m;
bool st[N];
int dmin[N], dmax[N];
int q[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(){
    
    
    cin >> 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);
        // 只能写成 add  ... ;  if (c == 2) 的形式
        // 如果写成 if (c == 1) ... ; else 会出错, 因此 c == 2是双向边, 上面这种写法, 少建了正向边
        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;
}

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转载自blog.csdn.net/esjiang/article/details/114357166