题意: n岛m条路,然后是 n个岛的坐标,然后是m条双向路,包括 岛和 岛 之间 最大客流量,让求 最左边的岛 到右边的岛 最大客流量
分析: 建图 以 左边的岛为原点,最右边的为终点求最大客流量。
刘汝佳STL会超时,因为vector牵扯到 复制操作
然后看大神的将stl全换了 http://www.cnblogs.com/xiao-xin/articles/4452306.html
1 #include <iostream> 2 #include <cstring> 3 #include <vector> 4 #include <algorithm> 5 #include <cstdio> 6 #include <queue> 7 using namespace std; 8 const int INF = 0x3f3f3f3f; 9 const int Max = 100000 + 10; 10 struct Edge 11 { 12 int to, flow, Next; 13 }; 14 struct Dinic 15 { 16 int n, s, t, now; 17 Edge edges[Max * 2]; 18 bool vis[Max]; 19 int d[Max], head[Max], q[Max]; 20 void init(int _s, int _t, int _n) 21 { 22 s = _s; t = _t; n = _n; 23 memset(head, -1, sizeof(head)); 24 now = 0; 25 } 26 void addEdge(int from, int to, int flow) 27 { 28 edges[now].to = to; edges[now].flow = flow; 29 edges[now].Next = head[from]; head[from] = now++; 30 31 edges[now].to = from; edges[now].flow = flow; 32 edges[now].Next = head[to]; head[to] = now++; 33 } 34 bool BFS() 35 { 36 memset(vis, 0, sizeof(vis)); 37 int Front =0, rear = 0; 38 q[rear++] = s; 39 d[s] = 0; 40 vis[s] = 1; 41 while (Front < rear) 42 { 43 int x = q[Front++]; 44 for (int i = head[x]; i != -1; i = edges[i].Next) 45 { 46 if ( vis[edges[i].to] == 0 && edges[i].flow > 0) 47 { 48 vis[edges[i].to] = 1; 49 d[ edges[i].to ] = d[x] + 1; 50 q[ rear++ ] = edges[i].to; 51 } 52 } 53 } 54 return vis[t]; 55 } 56 int DFS(int x, int a) 57 { 58 if (x == t || a == 0) 59 return a; 60 int flow = 0, f; 61 for (int i = head[x]; i != -1; i = edges[i].Next) 62 { 63 if (d[x] + 1 == d[edges[i].to] && edges[i].flow > 0) 64 { 65 f = DFS(edges[i].to, min(a - flow, edges[i].flow)); 66 edges[i].flow -= f; 67 edges[i ^ 1].flow += f; 68 flow += f; 69 if (flow == a) 70 return flow; 71 } 72 } 73 if (flow == 0) 74 d[x] = 0; 75 return flow; 76 } 77 int MaxFlow() 78 { 79 int flow = 0; 80 while ( BFS() ) 81 { 82 flow += DFS(s, INF); 83 } 84 return flow; 85 } 86 87 }g; 88 /* 89 struct Dinic 90 { 91 int n, s, t; 92 vector<Edge> edges; 93 vector<int> G[Max]; 94 bool vis[Max]; 95 int d[Max]; 96 int cur[Max]; 97 void init(int _n, int _s, int _t) 98 { 99 n = _n; 100 s = _s; 101 t = _t; 102 edges.clear(); 103 for (int i = 0; i <= n; i++) 104 G[i].clear(); 105 } 106 void addEdge(int from, int to, int cap) 107 { 108 Edge edge; 109 edge.from = from; 110 edge.to = to; 111 edge.cap = cap; 112 edge.flow = 0; 113 edges.push_back(edge); 114 edge.from = to; 115 edge.to = from; 116 edge.cap = cap; 117 edge.flow = 0; 118 edges.push_back(edge); 119 int m = (int) edges.size(); 120 G[from].push_back(m - 2); 121 G[to].push_back(m - 1); 122 } 123 bool BFS() 124 { 125 memset(vis, 0, sizeof(vis)); 126 queue<int> Q; 127 Q.push(s); 128 d[s] = 0; 129 vis[s] = 1; 130 while (!Q.empty()) 131 { 132 int x = Q.front(); 133 Q.pop(); 134 for (int i = 0; i < (int) G[x].size(); i++) 135 { 136 Edge & e = edges[ G[x][i] ]; 137 if (!vis[e.to] && e.cap > e.flow) 138 { 139 vis[e.to] = 1; 140 d[e.to] = d[x] + 1; 141 Q.push(e.to); 142 } 143 } 144 } 145 return vis[t]; 146 } 147 int DFS(int x, int a) 148 { 149 if (x == t || a == 0) 150 return a; 151 int flow = 0, f; 152 for (int& i = cur[x]; i < (int)G[x].size(); i++) 153 { 154 Edge & e = edges[ G[x][i] ]; 155 if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) 156 { 157 e.flow += f; 158 edges[G[x][i] ^ 1].flow -= f; 159 flow += f; 160 a -= f; 161 if (a == 0) 162 break; 163 } 164 } 165 return flow; 166 } 167 int MaxFlow() 168 { 169 int flow = 0; 170 while( BFS() ) 171 { 172 memset(cur, 0, sizeof(cur)); 173 flow += DFS(s, INF); 174 } 175 return flow; 176 } 177 178 179 }g; 180 */ 181 int main() 182 { 183 int T, n, m, east, west; 184 scanf("%d", &T); 185 while (T--) 186 { 187 scanf("%d%d", &n, &m); 188 int maxx = -INF, minx = INF; 189 int x, y, w; 190 for (int i = 1; i <= n; i++) 191 { 192 scanf("%d%d", &x, &y); 193 if (x > maxx) 194 { 195 maxx = x; 196 west = i; 197 } 198 if (x < minx) 199 { 200 minx = x; 201 east = i; 202 } 203 } 204 g.init(east, west, n); 205 for (int i = 1; i <= m; i++) 206 { 207 scanf("%d%d%d", &x, &y, &w); 208 g.addEdge(x, y, w); 209 } 210 printf("%d ", g.MaxFlow()); 211 } 212 return 0; 213 }