hdu5889（最大流+最短路）

传送门

题解：给一幅图，在这幅图的最短路上求最大流。

先跑一边最短路，然后将最短路上的边建图，跑一边最大流就好了。
建图我是用dist[u] + 1 ==dist[v]来判断他是不是最短路上的边的。

（弱智WA了一发。强行另存了一次边，然后强制将最短路上点标号小的向点标号大的建边。）

#include<bits/stdc++.h>
using namespace std;
const int MAXN=10010;
const int INF=0x3f3f3f3f;
const int MAXM = 400010;
struct Edge
{
    int v;
    int cost;
    Edge(int _v=0,int _cost=0):v(_v),cost(_cost) {}
};
vector<Edge> E[MAXN];
void addedge(int u,int v,int w)
{
    E[u].push_back(Edge(v,w));
}
bool vis[MAXN];//在队列标志
int cnt[MAXN];//每个点的入队列次数
int dist[MAXN];
int v,u,w;
void SPFA(int start,int n)
{
    memset(vis,false,sizeof(vis));
    for(int i=1; i<=n; i++)dist[i]=INF;
    vis[start]=true;
    dist[start]=0;
    queue<int>que;
    while(!que.empty())que.pop();
    que.push(start);
    memset(cnt,0,sizeof(cnt));
    cnt[start]=1;
    while(!que.empty())
    {
        int u=que.front();
        que.pop();
        vis[u]=false;
        for(int i=0; i<E[u].size(); i++)
        {
            int v=E[u][i].v;
            if(dist[v]>dist[u]+1)
            {
                dist[v]=dist[u]+1;
                if(!vis[v])
                {
                    vis[v]=true;
                    que.push(v);
                    if(++cnt[v]>n)return ;
//cnt[i]为入队列次数，用来判定是否存在负环回路
                }
            }
        }
    }
    return ;
}
struct Edges
{
    int to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init()
{
    tol = 0;
    memset(head,-1,sizeof(head));
}
//加边，单向图三个参数，双向图四个参数
void add(int u,int v,int w,int rw=0)
{
    edge[tol].to = v;
    edge[tol].cap = w;
    edge[tol].next = head[u];
    edge[tol].flow = 0;
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = rw;
    edge[tol].next = head[v];
    edge[tol].flow = 0;
    head[v]=tol++;
}
int sap(int start,int end,int N)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            for(int i = pre[u]; i != -1; i = pre[edge[i^1].to])
                if(Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            for(int i = pre[u]; i != -1; i = pre[edge[i^1].to])
            {
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        bool flag = false;
        int v;
        for(int i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
            {
                flag = true;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if(flag)
        {
            u = v;
            continue;
        }
        int Min = N;
        for(int i = head[u]; i != -1; i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if(!gap[dep[u]])return ans;
        dep[u] = Min+1;
        gap[dep[u]]++;
        if(u != start) u = edge[pre[u]^1].to;
    }
    return ans;
}
int main()
{
    int t,n,m;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d %d",&n,&m);
        init();
        for (int i = 1; i <= n; i++)
        {
            E[i].clear();
        }
        for (int i = 0; i < m; i++)
        {
            scanf("%d %d %d",&u,&v,&w);
            addedge(u,v,w);
            addedge(v,u,w);
        }
        SPFA(1,n);
        for (int i = 1; i <= n; i++)
        {
            for(int j = 0; j < E[i].size(); j++)
            if(dist[i] + 1 == dist[E[i][j].v])
            {
                add(i,E[i][j].v,E[i][j].cost);
            }
        }
        printf("%d\n",sap(1,n,n));
    }
}