[CQOI2015] 网络吞吐量 - 最大流,最短路

标签：make   +=   second   str   inter   情况   情况下   clear   cout   

在第i个点只能选A[i]次的情况下,能选出多少条1-n的最短路

Solution

我们造出最短路DAG，然后对每个点拆点限流，跑最大流即可

双向边警告！（有悖直觉

#include <bits/stdc++.h>
using namespace std;
#define int long long
namespace sp {
    // Interfaces: n, g[][], spt[], d[]
    const int N = 200005;
    int n;
    struct edge {int u,v,w;};
    vector<pair<int, int> >g[N];
    vector<edge> spt;
    int d[N], v[N];
    void make(int t1, int t2, int t3) {
        g[t1].push_back(make_pair(t2, t3));
    }
    void reset_graph() {
        for (int i = 0; i <= n; i++)
            g[i].clear();
    }
    void reset_solver() {
        memset(d, 0x3f, sizeof d);
        memset(v, 0x00, sizeof v);
    }
    void solve(int v0) {
        priority_queue<pair<int,int>,vector<pair<int,int> >,greater<pair<int,int> > >q;
        reset_solver();
        d[v0] = 0;
        q.push(make_pair(0,v0));
        while (q.size()) {
            pair<int, int> p = q.top();
            int dis = p.first;
            int pos = p.second;
            q.pop();
            v[pos] = 1;
            for (int i = 0; i < g[pos].size(); i++) {
                int x = g[pos][i].first;
                int y = g[pos][i].second;
                if (d[x] > d[pos] + y) {
                    d[x] = d[pos] + y;
                    if (!v[x]) q.push(make_pair(d[x], x));
                }
            }
        }
        for(int i=1;i<=n;i++) {
            int p=i;
            for(int j=0;j<g[p].size();j++) {
                int q=g[p][j].first, w=g[p][j].second;
                if(d[q]==d[p]+w) spt.push_back((edge){p,q,w});
            }
        }
    }
}


namespace flow {

const int maxn = 200005;
const int inf = 1e+14;

int dis[maxn], ans, cnt = 1, s, t, pre[maxn * 10], nxt[maxn * 10], h[maxn], v[maxn * 10];
std::queue<int> q;
void make(int x, int y, int z) {
    pre[++cnt] = y, nxt[cnt] = h[x], h[x] = cnt, v[cnt] = z;
    pre[++cnt] = x, nxt[cnt] = h[y], h[y] = cnt;
}
bool bfs() {
    memset(dis, 0, sizeof dis);
    q.push(s), dis[s] = 1;
    while (!q.empty()) {
        int x = q.front();
        q.pop();
        for (int i = h[x]; i; i = nxt[i])
            if (!dis[pre[i]] && v[i])
                dis[pre[i]] = dis[x] + 1, q.push(pre[i]);
    }
    return dis[t];
}
int dfs(int x, int flow) {
    if (x == t || !flow)
        return flow;
    int f = flow;
    for (int i = h[x]; i; i = nxt[i])
        if (v[i] && dis[pre[i]] > dis[x]) {
            int y = dfs(pre[i], min(v[i], f));
            f -= y, v[i] -= y, v[i ^ 1] += y;
            if (!f)
                return flow;
        }
    if (f == flow)
        dis[x] = -1;
    return flow - f;
}
int solve(int _s,int _t) {
    s=_s;
    t=_t;
    ans = 0;
    for (; bfs(); ans += dfs(s, inf));
    return ans;
}
}

const int N = 5005;
int n,m,t1,t2,t3,lim[N];

signed main() {
    scanf("%lld%lld",&n,&m);
    sp::n=n;
    for(int i=1;i<=m;i++) {
        scanf("%lld%lld%lld",&t1,&t2,&t3);
        sp::make(t1,t2,t3);
        sp::make(t2,t1,t3);
    }
    sp::solve(1);
    //for(int i=1;i<=n;i++) cout<<sp::d[i]<<" ";
    //cout<<endl;
    for(int i=0;i<sp::spt.size();i++) {
        int u=sp::spt[i].u, v=sp::spt[i].v, w=sp::spt[i].w;
        flow::make(2*u,2*v-1,1e+14);
    }
    for(int i=1;i<=n;i++) {
        scanf("%lld",&lim[i]);
        flow::make(2*i-1,2*i,lim[i]);
    }
    cout<<flow::solve(2,2*n-1);
}

我的模板丑的不行了，我得去改造模板了

[CQOI2015] 网络吞吐量 - 最大流,最短路

标签：make   +=   second   str   inter   情况   情况下   clear   cout   

原文地址：https://www.cnblogs.com/mollnn/p/12286351.html