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。
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1 // File Name: 2391.cpp 2 // Author: Missa 3 // Created Time: 2013/4/17 星期三 0:02:23 4 5 #include<iostream> 6 #include<cstdio> 7 #include<cstring> 8 #include<algorithm> 9 #include<cmath> 10 #include<queue> 11 #include<stack> 12 #include<string> 13 #include<vector> 14 #include<cstdlib> 15 #include<map> 16 #include<set> 17 using namespace std; 18 #define CL(x,v) memset(x,v,sizeof(x)); 19 #define R(i,st,en) for(int i=st;i<en;++i) 20 #define LL long long 21 22 const LL INF = 1e16; 23 const int inf = 0x3f3f3f3f; 24 const int maxn = 410; 25 const int maxm = 160010; 26 struct Edge 27 { 28 int v, c, next;//指向点,流量,下一个点 29 }p[maxm << 1]; 30 int head[maxn], e; 31 int d[maxn], cur[maxn];//层次记录，当前弧优化 32 int n, m, st, en; 33 void init() 34 { 35 e = 0; 36 memset(head, -1, sizeof(head)); 37 } 38 void addEdge(int u, int v, int c) 39 { 40 p[e].v = v; p[e].c = c; 41 p[e].next = head[u];head[u] = e++; 42 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. 43 p[e].next = head[v];head[v] = e++; 44 } 45 int bfs(int st, int en) 46 { 47 queue <int> q; 48 memset(d, 0, sizeof(d)); 49 d[st] = 1; 50 q.push(st); 51 while (!q.empty()) 52 { 53 int u = q.front(); q.pop(); 54 for (int i = head[u]; i != -1; i = p[i].next) 55 { 56 if (p[i].c > 0 && !d[p[i].v]) 57 { 58 d[p[i].v] = d[u] + 1; 59 q.push(p[i].v); 60 } 61 } 62 } 63 return d[en]; 64 } 65 int dfs(int u, int a)//a表示流量 66 { 67 if (u == en || a == 0) return a; 68 int f, flow = 0; 69 for (int& i = cur[u]; i != -1; i = p[i].next) 70 { 71 if (d[u] + 1 == d[p[i].v] && (f = dfs(p[i].v,min(a,p[i].c))) > 0) 72 { 73 p[i].c -= f; 74 p[i^1].c += f; 75 flow += f; 76 a -= f; 77 if (!a) break; 78 } 79 } 80 return flow; 81 } 82 int dinic(int st, int en) 83 { 84 int res = 0; 85 while (bfs(st, en)) 86 { 87 for (int i = 0; i <= n; ++i) 88 cur[i] = head[i]; 89 res += dfs(st, inf); 90 } 91 return res; 92 } 93 int w[maxn], c[maxn], tt; 94 LL a[maxn][maxn]; 95 void rebuild(LL mid) 96 { 97 init(); 98 for (int i = 1; i <= tt; ++i) 99 { 100 addEdge(st, i, w[i]); 101 addEdge(i + tt, en, c[i]); 102 } 103 for (int i = 1; i <= tt; ++i) 104 addEdge(i, i + tt, inf); 105 for (int i = 1; i <= tt; ++i) 106 { 107 for (int j = i + 1; j <= tt; ++j) 108 if (a[i][j] <= mid) 109 { 110 addEdge(i, j + tt, inf); 111 addEdge(j, i + tt, inf); 112 } 113 } 114 } 115 int main() 116 { 117 while(~scanf("%d%d",&n,&m)) 118 { 119 tt = n; 120 st = 0; en = (n << 1) + 1; 121 int sum = 0; 122 for (int i = 1; i <= n; ++i) 123 { 124 scanf("%d%d",&w[i], &c[i]); 125 sum += w[i]; 126 } 127 for (int i = 1; i <= n; ++i) 128 for (int j = 1; j <= n; ++j) 129 a[i][j] = INF; 130 //cout <<a[n][n] + a[n][n]<<endl; 131 for (int i = 1; i <= m; ++i) 132 { 133 int uu, vv, len; 134 scanf("%d%d%d", &uu, &vv, &len); 135 if (len < a[uu][vv]) 136 a[uu][vv] = a[vv][uu] = len; 137 } 138 for (int k = 1; k <= n; ++k) 139 for (int i = 1; i <= n; ++i) 140 for (int j = 1; j <= n; ++j) 141 if (a[i][j] > a[i][k] + a[k][j]) 142 a[i][j] = a[i][k] + a[k][j]; 143 LL ans = -1, low = 0, high = INF - 1; 144 n = en; 145 while (low <= high) 146 { 147 LL mid = (low + high) / 2; 148 //cout << mid <<endl; 149 rebuild(mid); 150 int tmp = dinic(st, en); 151 //cout <<"mid = "<<mid<<" dinic = "<< tmp <<endl; 152 if (tmp >= sum) 153 { 154 ans = mid; 155 high = mid - 1; 156 } 157 else low = mid + 1; 158 } 159 printf("%I64d\n",ans); 160 } 161 return 0; 162 }