poj 2391

无向图中给出N个点，M条边，每个点有当前囤积流量和最大容量，任意两个点之间有距离，首先保证所有的流量都可以被节点吸收（即不超过最大容量），然后将流量跑的最大距离限制到最小。

拆点：

然后二分答案。跟poj 2112差不多。

// File Name: 2391.cpp
// Author: Missa
// Created Time: 2013/4/17 星期三 0:02:23

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<vector>
#include<cstdlib>
#include<map>
#include<set>
using namespace std;
#define CL(x,v) memset(x,v,sizeof(x));
#define R(i,st,en) for(int i=st;i<en;++i)
#define LL long long

const LL INF = 1e16;
const int inf = 0x3f3f3f3f;
const int maxn = 410;
const int maxm = 160010;
struct Edge
{
    int v, c, next;//指向点,流量,下一个点
}p[maxm << 1];
int head[maxn], e;
int d[maxn], cur[maxn];//层次记录，当前弧优化
int n, m, st, en;
void init()
{
    e = 0;
    memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int c)
{
    p[e].v = v; p[e].c = c; 
    p[e].next = head[u];head[u] = e++;
    p[e].v = u; p[e].c = 0; //无向图时逆边赋值流量cap，有向图时赋值0.
    p[e].next = head[v];head[v] = e++;
}
int bfs(int st, int en)
{
    queue <int> q;
    memset(d, 0, sizeof(d));
    d[st] = 1;
    q.push(st);
    while (!q.empty())
    {
        int u = q.front(); q.pop();
        for (int i = head[u]; i != -1; i = p[i].next)
        {
            if (p[i].c > 0 && !d[p[i].v])
            {
                d[p[i].v] = d[u] + 1;
                q.push(p[i].v);
            }
        }
    }
    return d[en];
}
int dfs(int u, int a)//a表示流量
{
    if (u == en || a == 0) return a;
    int f, flow = 0;
    for (int& i = cur[u]; i != -1; i = p[i].next)
    {
        if (d[u] + 1 == d[p[i].v] && (f = dfs(p[i].v,min(a,p[i].c))) > 0)
        {
            p[i].c -= f;
            p[i^1].c += f;
            flow += f;
            a -= f;
            if (!a) break;
        }
    }
    return flow;
}
int dinic(int st, int en)
{
    int res = 0;
    while (bfs(st, en))
    {
        for (int i = 0; i <= n; ++i)
            cur[i] = head[i];
        res += dfs(st, inf);
    }
    return res;
}
int w[maxn], c[maxn], tt; 
LL a[maxn][maxn];
void rebuild(LL mid)
{
    init();
    for (int i = 1; i <= tt; ++i)
    {
        addEdge(st, i, w[i]);
        addEdge(i + tt, en, c[i]);
    }
    for (int i = 1; i <= tt; ++i)
        addEdge(i, i + tt, inf);
    for (int i = 1; i <= tt; ++i)
    {
        for (int j = i + 1; j <= tt; ++j)
            if (a[i][j] <= mid)
            {
                addEdge(i, j + tt, inf);
                addEdge(j, i + tt, inf);
            }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        tt = n;
        st = 0; en = (n << 1) + 1;
        int sum = 0;
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d%d",&w[i], &c[i]);
            sum += w[i];
        }
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                a[i][j] = INF;
        //cout <<a[n][n] + a[n][n]<<endl;
        for (int i = 1; i <= m; ++i)
        {
            int uu, vv, len;
            scanf("%d%d%d", &uu, &vv, &len);
            if (len < a[uu][vv])
                a[uu][vv] = a[vv][uu] = len;
        }
        for (int k = 1; k <= n; ++k)
            for (int i = 1; i <= n; ++i)
                for (int j = 1; j <= n; ++j)
                    if (a[i][j]  > a[i][k] + a[k][j])
                        a[i][j] = a[i][k] + a[k][j];
        LL ans = -1, low = 0, high = INF - 1;
        n = en;
        while (low <= high)
        {
            LL mid = (low + high) / 2;
            //cout << mid <<endl;
            rebuild(mid);
            int tmp = dinic(st, en);
            //cout <<"mid = "<<mid<<" dinic = "<< tmp <<endl;
            if (tmp >= sum)
            {
                ans = mid;
                high = mid - 1;
            }
            else low = mid + 1;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}