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; // vertexint cap; // capacityint flow; // current flow in this arcint 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0edge[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;
}