POJ 2391 拆点+最大流+二分

题意就不再说了，主要是想说一下为啥要拆点

POJ 2112跟本题很相似，也是二分，但它就没用拆点，本题就用了

为啥呢？ 因为POJ2112上建的图中是一个很明显的二分图，也就是左边的点绝对不会跟左边的点去连边的。而本题中就不一样了，任何点之间都可能有边。

会出现这种情况，1->2->3，这是一条链，而1->2最短路为1，2->3为1，1->3为 2，此时如果我们限制最大距离为1的话，建的新图中必然有1->2,2-3, 然后我们就发现问题了，新图中1和3也会间接的连接起来，而我们显然是不想这么让它流的。这就需要拆点了，对每个点x，拆成x'和x''，然后x'和x''之间有一条无限容量的边，这样的话，随便多少牛路过这个点都是可以的，如果i->j这条边符合了距离限制，就加i'->j''  所有的边加完后，建立源点，对所有的x'连边，容量为已经有的牛，汇点的话，就用所有的j''连接汇点，容量就是能容纳的牛的数量。

这样一拆点，就发现之前的问题不复存在了，还是比如1->2->3这个例子，加的边是1’->2''，2'->3'' 不会有流从1流到3去，因为加的每条边都流向了汇点

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define MAXN 555
#define MAXM 222222
#define INF 1000000007
using namespace std;
struct node
{
    int ver;    // vertex
    int cap;    // capacity
    int flow;   // current flow in this arc
    int next, rev;
}edge[MAXM];
int dist[MAXN], numbs[MAXN], src, des, n;
int head[MAXN], e;
void add(int x, int y, int c)
{       //e记录边的总数
    edge[e].ver = y;
    edge[e].cap = c;
    edge[e].flow = 0;
    edge[e].rev = e + 1;        //反向边在edge中的下标位置
    edge[e].next = head[x];   //记录以x为起点的上一条边在edge中的下标位置
    head[x] = e++;           //以x为起点的边的位置
    //反向边
    edge[e].ver = x;
    edge[e].cap = 0;  //反向边的初始网络流为0
    edge[e].flow = 0;
    edge[e].rev = e - 1;
    edge[e].next = head[y];
    head[y] = e++;
}
void rev_BFS()
{
    int Q[MAXN], qhead = 0, qtail = 0;
    for(int i = 1; i <= n; ++i)
    {
        dist[i] = MAXN;
        numbs[i] = 0;
    }
    Q[qtail++] = des;
    dist[des] = 0;
    numbs[0] = 1;
    while(qhead != qtail)
    {
        int v = Q[qhead++];
        for(int i = head[v]; i != -1; i = edge[i].next)
        {
            if(edge[edge[i].rev].cap == 0 || dist[edge[i].ver] < MAXN)continue;
            dist[edge[i].ver] = dist[v] + 1;
            ++numbs[dist[edge[i].ver]];
            Q[qtail++] = edge[i].ver;
        }
    }
}
void init()
{
    e = 0;
    memset(head, -1, sizeof(head));
}
int maxflow()
{
    int u, totalflow = 0;
    int Curhead[MAXN], revpath[MAXN];
    for(int i = 1; i <= n; ++i)Curhead[i] = head[i];
    u = src;
    while(dist[src] < n)
    {
        if(u == des)     // find an augmenting path
        {
            int augflow = INF;
            for(int i = src; i != des; i = edge[Curhead[i]].ver)
                augflow = min(augflow, edge[Curhead[i]].cap);
            for(int i = src; i != des; i = edge[Curhead[i]].ver)
            {
                edge[Curhead[i]].cap -= augflow;
                edge[edge[Curhead[i]].rev].cap += augflow;
                edge[Curhead[i]].flow += augflow;
                edge[edge[Curhead[i]].rev].flow -= augflow;
            }
            totalflow += augflow;
            u = src;
        }
        int i;
        for(i = Curhead[u]; i != -1; i = edge[i].next)
            if(edge[i].cap > 0 && dist[u] == dist[edge[i].ver] + 1)break;
        if(i != -1)     // find an admissible arc, then Advance
        {
            Curhead[u] = i;
            revpath[edge[i].ver] = edge[i].rev;
            u = edge[i].ver;
        }
        else        // no admissible arc, then relabel this vertex
        {
            if(0 == (--numbs[dist[u]]))break;    // GAP cut, Important!
            Curhead[u] = head[u];
            int mindist = n;
            for(int j = head[u]; j != -1; j = edge[j].next)
                if(edge[j].cap > 0)mindist = min(mindist, dist[edge[j].ver]);
            dist[u] = mindist + 1;
            ++numbs[dist[u]];
            if(u != src)
                u = edge[revpath[u]].ver;    // Backtrack
        }
    }
    return totalflow;
}
int F, P;
int now[MAXN], can[MAXN];
long long d[MAXN][MAXN];
void floyd(int n)
{
    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                if(d[i][k] + d[k][j] < d[i][j])
                    d[i][j] = d[i][k] + d[k][j];
}
int main()
{
    int total = 0;
    scanf("%d%d", &F, &P);
    for(int i = 1; i <= F; i++)
    {
        scanf("%d%d", &now[i], &can[i]);
        total += now[i];
    }
    for(int i = 1; i <= F; i++)
        for(int j = 1; j <= F; j++)
            if(i != j) d[i][j] = 2000000000000LL;
            else d[i][j] = 0;
    int u, v, w;
    for(int i = 1; i <= P; i++)
    {
        scanf("%d%d%d", &u, &v, &w);
        if(d[u][v] > w) d[u][v] = d[v][u] = w;
    }
    floyd(F);
    long long low = 0, high = 0;
    for(int i = 1; i <= F; i++)
        for(int j = 1; j <= F; j++)
            if(d[i][j] != 2000000000000LL) high = max(high, d[i][j]);
    n = 2 * F + 2;
    src = 1;
    des = n;
    long long ans = -1;
    while(low <= high)
    {
        long long mid = (low + high) / 2;
        init();
        for(int i = 1; i <= F; i++)
            for(int j = 1; j <= F; j++)
                if(d[i][j] <= mid)
                    add(i + 1, j + 1 + F, INF);
        for(int i = 1; i <= F; i++)
        {
            add(src, i + 1, now[i]);
            add(i + 1 + F, des, can[i]);
        }
        rev_BFS();
        int flow = maxflow();
        //printf("dsfsd %d\n", flow);
        if(flow == total)
        {
            ans = mid;
            high = mid - 1;
        }
        else low = mid + 1;
    }
    printf("%lld\n", ans);
    return 0;
}