题目大意：
给出一张无向图，找出T条从1..N的路径，互不重复，求走过的所有边中的最大值最小是多少。
算法讨论：
首先最大值最小就提醒我们用二分，每次二分一个最大值，然后重新构图，把那些边权符合要求的边加入新图，在新图上跑网络流。
这题有许多注意的地方：
1、因为是无向图，所以我们在加正向边和反向边的时候，流量都是1，而不是正向边是1，反向边是0。
2、题目中说这样的路径可能不止t条，所以我们在最后二分判定的时候不能写 == t，而是要 >= t。
3、这题很容易T ，表示我T了N遍，弱菜啊。
Codes：
1 #include <cstring> 2 #include <cstdlib> 3 #include <iostream> 4 #include <algorithm> 5 #include <cstdio> 6 #include <vector> 7 #include <queue> 8 9 using namespace std; 10 11 struct Edge{ 12 int from, to, cap, flow; 13 Edge(int _from = 0, int _to = 0, int _cap = 0, int _flow = 0): 14 from(_from), to(_to), cap(_cap), flow(_flow) {} 15 }E[40000 + 5]; 16 17 struct Dinic{ 18 static const int N = 200 + 5; 19 static const int M = 80000 + 5; 20 static const int oo = 0x3f3f3f3f; 21 22 int n, m, s, t; 23 vector <Edge> edges; 24 vector <int> G[N]; 25 int dis[N], cur[N]; 26 bool vi[N]; 27 28 void Clear(){ 29 for(int i = 1; i <= n; ++ i) G[i].clear(); 30 edges.clear(); 31 } 32 33 void Add(int from, int to, int cap, int flow){ 34 edges.push_back((Edge){from, to, cap, 0}); 35 edges.push_back((Edge){to, from, cap, 0}); 36 int m = edges.size(); 37 G[from].push_back(m - 2); 38 G[to].push_back(m - 1); 39 } 40 41 bool bfs(){ 42 for(int i = 1; i <= n; ++ i) vi[i] = false; 43 dis[s] = 0; vi[s] = true; 44 queue <int> q; q.push(s); 45 46 while(!q.empty()){ 47 int x = q.front(); q.pop(); 48 for(int i = 0; i < G[x].size(); ++ i){ 49 Edge &e = edges[G[x][i]]; 50 if(!vi[e.to] && e.cap > e.flow){ 51 vi[e.to] = true; 52 dis[e.to] = dis[x] + 1; 53 q.push(e.to); 54 } 55 } 56 } 57 return vi[t]; 58 } 59 60 int dfs(int x, int a){ 61 if(x == t || a == 0) return a; 62 int flw = 0, f; 63 for(int &i = cur[x]; i < G[x].size(); ++ i){ 64 Edge &e = edges[G[x][i]]; 65 if(dis[x] + 1 == dis[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0){ 66 e.flow += f; edges[G[x][i]^1].flow -= f; 67 a -= f; flw += f; 68 if(!a) break; 69 } 70 } 71 return flw; 72 } 73 74 int Maxflow(int s, int t){ 75 this->s = s;this-> t = t; 76 int flw = 0; 77 while(bfs()){ 78 memset(cur, 0, sizeof cur); 79 flw += dfs(s, oo); 80 } 81 return flw; 82 } 83 84 }Net; 85 86 int ns, ps, ts; 87 int l = 0x3f3f3f3f, r, mid; 88 89 bool check(int mv){ 90 for(int i = 0; i < ps; ++ i){ 91 if(E[i].cap <= mv){ 92 Net.Add(E[i].from, E[i].to, 1, 0); 93 } 94 } 95 return Net.Maxflow(1, Net.n) >= ts; 96 } 97 98 int Solve(){ 99 int ans; 100 101 while(l <= r){ 102 Net.Clear(); 103 mid = l + (r - l) / 2; 104 if(check(mid)){ 105 ans = mid; r = mid - 1; 106 } 107 else l = mid + 1; 108 } 109 110 return ans; 111 } 112 113 int main(){ 114 int x, y, z; 115 116 scanf("%d%d%d", &ns, &ps, &ts); 117 Net.n = ns; 118 for(int i = 0; i < ps; ++ i){ 119 scanf("%d%d%d", &x, &y, &z); 120 E[i] = (Edge){x, y, z, 0}; 121 l = min(l, z); r = max(r, z); 122 } 123 124 printf("%d ", Solve()); 125 return 0; 126 }
让人更加不能理解的是，为什么我换了上面的邻接表的形式，用STL容器来存邻接表就AC，用数组来存就T成狗。
下面是我狂T的代码，良心网友们给查查错误呗。
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 #include <queue> 7 8 using namespace std; 9 10 int ns, ps, ts, te; 11 int l=0x3f3f3f3f, r, Mid; 12 13 struct Edge{ 14 int from, to, dt; 15 }e[40000 + 5]; 16 17 struct Dinic{ 18 static const int maxn = 300 + 5; 19 static const int maxm = 80000 + 5; 20 static const int oo = 0x3f3f3f3f; 21 22 int n,m,s,t; 23 int tot; 24 int first[maxn],next[maxm]; 25 int u[maxm],v[maxm],cap[maxm],flow[maxm]; 26 int cur[maxn],dis[maxn]; 27 bool vi[maxn]; 28 29 Dinic(){tot=0;memset(first,-1,sizeof first);} 30 void Clear(){tot = 0;memset(first,-1,sizeof first);} 31 void Add(int from,int to,int cp,int flw){ 32 u[tot] = from;v[tot] = to;cap[tot] = cp;flow[tot] = 0; 33 next[tot] = first[u[tot]]; 34 first[u[tot]] = tot; 35 ++ tot; 36 } 37 bool bfs(){ 38 memset(vi,false,sizeof vi); 39 queue <int> q; 40 dis[s] = 0;vi[s] = true; 41 q.push(s); 42 43 while(!q.empty()){ 44 int now = q.front();q.pop(); 45 for(int i = first[now];i != -1;i = next[i]){ 46 if(!vi[v[i]] && cap[i] > flow[i]){ 47 vi[v[i]] = true; 48 dis[v[i]] = dis[now] + 1; 49 q.push(v[i]); 50 } 51 } 52 } 53 return vi[t]; 54 } 55 int dfs(int x,int a){ 56 if(x == t || a == 0) return a; 57 int flw=0,f; 58 int &i = cur[x]; 59 for(i = first[x];i != -1;i = next[i]){ 60 if(dis[x] + 1 == dis[v[i]] && (f = dfs(v[i],min(a,cap[i]-flow[i]))) > 0){ 61 flow[i] += f;flow[i^1] -= f; 62 a -= f;flw += f; 63 if(a == 0) break; 64 } 65 } 66 return flw; 67 } 68 int MaxFlow(int s,int t){ 69 this->s = s;this->t = t; 70 int flw=0; 71 while(bfs()){ 72 memset(cur,0,sizeof cur); 73 flw += dfs(s,oo); 74 } 75 return flw; 76 } 77 }Net; 78 79 bool check(int mid){ 80 Net.Clear(); 81 for(int i = 1; i <= ps; ++ i){ 82 if(e[i].dt <= mid){ 83 Net.Add(e[i].from, e[i].to, 1, 0); 84 Net.Add(e[i].to, e[i].from, 1, 0); 85 } 86 } 87 return Net.MaxFlow(1, Net.n) >= ts; 88 } 89 int Solve(){ 90 int ans; 91 while(l <= r){ 92 Mid = l + (r - l) / 2; 93 if(check(Mid)){ 94 ans = Mid; r = Mid - 1; 95 } 96 else l = Mid + 1; 97 } 98 return ans; 99 } 100 int main(){ 101 int x, y, z; 102 scanf("%d%d%d", &ns, &ps, &ts); 103 Net.n = ns;te = 0; 104 for(int i = 1; i <= ps; ++ i){ 105 scanf("%d%d%d", &x, &y, &z); 106 ++ te; 107 e[te].from = x;e[te].to = y;e[te].dt = z; 108 l = min(l, z); r = max(z, r); 109 } 110 printf("%d ", Solve()); 111 return 0; 112 }