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最大流问题求解(ISAP算法)

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 1e4 + 5, M = 2e5 + 5, INF = 1e8;
int h[N], e[M], ne[M], w[M], idx, d[N], q[N], n, m, S, T, cur[N], gap[N];
long long maxflow;
void add(int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
    e[idx] = a, ne[idx] = h[b], w[idx] = 0, h[b] = idx ++;
}

void bfs()
{
    int hh = 0, tt = 0;
    memset(d, -1, sizeof d);
    q[0] = T, d[T] = 0, cur[T] = h[T], gap[0] = 1;
    while (hh <= tt)
    {
        int t = q[hh ++ ];
        for (int i = h[t]; ~i; i = ne[i])
        {
            int ver = e[i];
            if (d[ver] == -1 && !w[i])
            {
                cur[ver] = h[ver];
                d[ver] = d[t] + 1;
                gap[d[ver]] ++;
                q[++ tt] = ver;
            }
        }
    }
}
int min(int a, long long b)
{
    return a < b ? a : b;
}
long long find(int u, long long limit)
{
    if (u == T)
    {
        maxflow += limit;
        return limit;
    }
    int flow = 0;

    for (int i = cur[u]; ~i; i = ne[i])
    {
        cur[u] = i;
        int ver = e[i];
        if (d[ver] + 1 == d[u] && w[i])
        {
            int t = find (ver, min(w[i], limit - flow));
            w[i] -= t, w[i ^ 1] += t, flow += t;
            if (flow == limit) return flow;
        }
    }
    -- gap[d[u]];
    if (gap[d[u]] == 0) d[S] = n + 1;
    d[u] ++;
    gap[d[u]] ++;
    return flow;
}

long long ISAP()
{
    bfs();
    while(d[S] < n)
    {
        memcpy(cur, h, sizeof h);
        find(S, INF);
    }
    return maxflow;
}

int main()
{
    scanf("%d%d%d%d", &n, &m, &S, &T);
    memset(h, -1, sizeof h);
    while (m -- )
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c);
    }

    printf("%lld\n", ISAP());

    return 0;
}

最小费用最大流

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 5010, M = 100010, INF = 1e8;

int n, m, S, T;
int h[N], e[M], f[M], w[M], ne[M], idx;
int q[N], d[N], pre[N], incf[N];
bool st[N];

void add(int a, int b, int c, int d)
{
    e[idx] = b, f[idx] = c, w[idx] = d, ne[idx] = h[a], h[a] = idx ++ ;
    e[idx] = a, f[idx] = 0, w[idx] = -d, ne[idx] = h[b], h[b] = idx ++ ;
}

bool spfa()
{
    int hh = 0, tt = 1;
    memset(d, 0x3f, sizeof d);
    memset(incf, 0, sizeof incf);
    q[0] = S, d[S] = 0, incf[S] = INF;
    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 ver = e[i];
            if (f[i] && d[ver] > d[t] + w[i])
            {
                d[ver] = d[t] + w[i];
                pre[ver] = i;
                incf[ver] = min(f[i], incf[t]);
                if (!st[ver])
                {
                    q[tt ++ ] = ver;
                    if (tt == N) tt = 0;
                    st[ver] = true;
                }
            }
        }
    }

    return incf[T] > 0;
}

void EK(int& flow, int& cost)
{
    flow = cost = 0;
    while (spfa())
    {
        int t = incf[T];
        flow += t, cost += t * d[T];
        for (int i = T; i != S; i = e[pre[i] ^ 1])
        {
            f[pre[i]] -= t;
            f[pre[i] ^ 1] += t;
        }
    }
}

int main()
{
    scanf("%d%d%d%d", &n, &m, &S, &T);
    memset(h, -1, sizeof h);
    while (m -- )
    {
        int a, b, c, d;
        scanf("%d%d%d%d", &a, &b, &c, &d);
        add(a, b, c, d);
    }

    int flow, cost;
    EK(flow, cost);
    printf("%d %d\n", flow, cost);

    return 0;
}

最短路径（堆优化）

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
typedef pair<int, int> PII;
const int N = 1.5e5 + 5;
int h[N], e[N], ne[N], w[N], idx, n, m;
int dis[N];
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

int dijkstra()
{
    memset(dis, 0x3f, sizeof dis);
    dis[1] = 0;
    priority_queue <PII, vector<PII>, greater<PII>> heap;
    heap.push({0, 1});
    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 != -1; i = ne[i])
        {
            int j = e[i];
            if (dis[j] > dis[ver] + w[i])
            {
                dis[j] = dis[ver] + w[i];
                heap.push({dis[j], j});
            }
        }
    }

    if (dis[n] == 0x3f3f3f3f) return -1;
    return dis[n];
}

int main()
{
    cin >> n >> m;
    memset(h, -1, sizeof h);
    while (m -- )
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }

    cout << dijkstra() << endl;
}

有边数限制的最短路

#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 510, M = 10010;

struct Edge
{
    int a, b, c;
}edges[M];

int n, m, k;
int dist[N];
int last[N];

void bellman_ford()
{
    memset(dist, 0x3f, sizeof dist);

    dist[1] = 0;
    for (int i = 0; i < k; i ++ )
    {
        memcpy(last, dist, sizeof dist);
        for (int j = 0; j < m; j ++ )
        {
            auto e = edges[j];
            dist[e.b] = min(dist[e.b], last[e.a] + e.c);
        }
    }
}

int main()
{
    scanf("%d%d%d", &n, &m, &k);

    for (int i = 0; i < m; i ++ )
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        edges[i] = {a, b, c};
    }

    bellman_ford();

    if (dist[n] > 0x3f3f3f3f / 2) puts("impossible");
    else printf("%d\n", dist[n]);

    return 0;
}

最小生成树

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 510, INF = 0x3f3f3f3f;
int g[N][N], dist[N], n, m;
bool st[N];
int prim()
{
    memset(dist, 0x3f, sizeof dist);
    int res = 0;
    for (int i = 0; i < n; i ++ )
    {
        int t = -1;
        for (int j = 1; j <= n; j ++ )
            if (!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;

        if (i && dist[t] == INF) return INF;

        if (i) res += dist[t];

        st[t] = true;

        for (int j = 1; j <= n; j ++ ) dist[j] = min(dist[j], g[t][j]);
    }
    return res;
}
int main()
{
    cin >> n >> m;
    memset(g, 0x3f, sizeof g);
    while (m -- )
    {
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = g[b][a] = min(g[a][b], c);
    }

    int t = prim();
    if (t == INF) puts("impossible");
    else cout << t << endl;
}

最短Hamilton路径

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 20, M = 1 << N;
int g[N][N], f[M][N];
int main()
{
    int n = 0;
    cin >> n;
    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < n; j ++ )
            cin >> g[i][j];

    memset(f, 0x3f, sizeof f);    
    f[1][0] = 0;
    for (int i = 0; i < 1 << n; i ++ )
        for (int j = 0; j < n; j ++ )
            if ((i >> j) & 1)
                for (int k = 0; k < n; k ++ )
                    if ((i >> k) & 1)
                        f[i][j] = min(f[i][j], f[i - (1 << j)][k] + g[k][j]);

    cout << f[(1 << n) - 1][n - 1] << endl;
}

欧拉回路

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 100100, M = 400100;

int h[N],e[M],ne[M],idx;
int ans[N*2],cnt;
bool used[M];
int din[N],dout[N];
int n,m,ver;

void add(int a,int b){
    e[idx] = b,ne[idx] = h[a],h[a] = idx++;
}

void dfs(int u){
    for(int &i = h[u]; ~i; ){
        if(used[i]){  //如果这条边用过了
            i = ne[i];   //删除这条边
            continue;
        }

        used[i] = true;  //标记这条边已使用
        if(ver == 1) used[i^1] = true;   //如果是无向图，那么这条边的反向边也要标记使用过了

        int t;
        if(ver == 1){
            t = i/2 + 1;
            if(i&1) t = -t;  //(0,1) (2,3) (4,5) 奇数编号是返回的边

        }else t = i+1;

        int j = e[i];
        i = ne[i];
        dfs(j);
        ans[cnt++] = t;
    }
}
int main()
{
    scanf("%d%d%d",&ver,&n,&m);
    memset(h,-1,sizeof h);

    for(int i = 0; i<m; i++){
        int a,b;
        scanf("%d%d",&a,&b);
        add(a,b);
        if(ver == 1) add(b,a);  //无向边
        din[b]++, dout[a]++;   
    }

    if(ver == 1){
        for(int i = 1; i<=n; i++){
            if(din[i]+dout[i] &1){
                //无向图含欧拉回路的充要条件是每个点的度都为偶数
                puts("NO");
                return 0;
            }
        }
    }else{
        for(int i = 1; i<=n; i++){
            if(din[i] != dout[i]){
                //有向图含欧拉回路的充要条件是每个点的入度等于出度
                puts("NO");
                return 0;
            }
        }
    }

    for(int i = 1; i<=n; i++){
        if(~h[i]) {
            dfs(i);
            break;
        }
    }

    if(cnt < m){
        puts("NO");
        return 0;
    }

    puts("YES");
    for(int i = cnt-1; i>=0; --i){
        cout<<ans[i]<<" ";
    }
    return 0;
}