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题意:
n个工厂加工电脑,电脑有p个部件。每个工厂加工的电脑必须是满足它对原材料的需求。p个部件,0代表不能有,1必须有,2代表可有可无。然后经过加工后,这个工厂输出的电脑各个部件的情况,0代表没有,1代表有。每个工厂有一个生产效率。求单位时间内的最多能生产的完整电脑数量。
解决:
每个工厂拆成两个点,流量为该工厂效率。超级源连接对p个部件需求都为0的工厂,超级汇连接输出p个部件都为1的工厂 。求一次最大流。
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #include <climits> 5 6 const int MAXN = 100 + 20; 7 const int INF = INT_MAX; 8 9 int g[MAXN]; 10 11 int n, p; 12 13 struct Mach{ 14 int q, in[11], out[11]; 15 }mach[MAXN<<2]; 16 17 struct Edge{ 18 int v, f, pre; 19 }e[MAXN*MAXN], e_bak[MAXN*MAXN]; 20 int e_n; 21 22 struct Flow{ 23 int u, v, f; 24 }flow[MAXN<<2]; 25 int f_e; 26 27 int layer[MAXN]; 28 29 int dfs(int u, int sink, int flow) 30 { 31 if(u == sink) 32 return flow; 33 int now_flow = 0; 34 for(int i = g[u]; i && flow; i = e[i].pre){ 35 int v = e[i].v; 36 int f = e[i].f; 37 if(f > 0 && layer[v] == layer[u] + 1){ 38 int d = dfs(v, sink, std::min(f, flow)); 39 e[i].f -= d; 40 e[i^1].f += d; 41 flow -= d; 42 now_flow += d; 43 } 44 } 45 return now_flow; 46 } 47 48 void bfs(int src, int sink) 49 { 50 memset(layer, -1, sizeof layer); 51 std::queue<int> que; 52 que.push(src); 53 layer[src] = 0; 54 while(que.empty() == false){ 55 int u = que.front(); 56 que.pop(); 57 for(int i = g[u]; i; i = e[i].pre){ 58 int v = e[i].v; 59 int f = e[i].f; 60 if(f > 0 && layer[v] == -1){ 61 que.push(v); 62 layer[v] = layer[u] + 1; 63 } 64 } 65 } 66 } 67 68 int dinic(int src, int sink) 69 { 70 int res = 0; 71 while(true){ 72 bfs(src, sink); 73 if(layer[sink] == -1){ 74 return res; 75 } 76 res += dfs(src, sink, INF); 77 } 78 } 79 80 81 82 void addEdge(int u, int v, int f) 83 { 84 ++ e_n; 85 e[e_n].v = v; 86 e[e_n].f = f; 87 e[e_n].pre = g[u]; 88 g[u] = e_n; 89 90 ++e_n; 91 e[e_n].v = u; 92 e[e_n].f = 0; 93 e[e_n].pre = g[v]; 94 g[v] = e_n; 95 } 96 97 void init() 98 { 99 e_n = 1; 100 f_e = 0; 101 memset(g, 0, sizeof g); 102 memset(e, 0, sizeof e); 103 memset(e_bak, 0, sizeof e_bak); 104 memset(flow, 0, sizeof flow); 105 memset(mach, 0, sizeof mach); 106 } 107 108 int main() 109 { 110 while(~scanf("%d%d", &p, &n)){ 111 init(); 112 //scanf("%d%d", &p, &n); 113 int src = 0; 114 int sink = n*2+1; 115 // input 116 for(int i = 1; i <= n; ++ i){ 117 scanf("%d", &mach[i].q); 118 for(int j = 1; j <= p; ++ j){ 119 scanf("%d", &mach[i].in[j]); 120 } 121 for(int j = 1; j <= p; ++ j){ 122 scanf("%d", &mach[i].out[j]); 123 } 124 addEdge(i, i+n, mach[i].q); 125 } 126 // addEdge 127 for(int i = 1; i <= n; ++ i){ 128 bool ok = true; 129 for(int j = 1; j <= p; ++ j){ 130 if(mach[i].in[j] == 1){ 131 ok = false; 132 break; 133 } 134 } 135 if(ok == true){ 136 addEdge(src, i, INF); 137 } 138 139 ok = true; 140 for(int j = 1; j <= p; ++ j){ 141 if(mach[i].out[j] != 1){ 142 ok = false; 143 break; 144 } 145 } 146 if(ok == true){ 147 addEdge(i+n, sink, INF); 148 } 149 150 for(int j = 1; j <= n; ++ j){ 151 if(i == j){ 152 continue; 153 } 154 ok = true; 155 for(int k = 1; k <= p; ++ k){ 156 if(mach[i].in[k] == 1 && mach[j].out[k] == 0){ 157 ok = false; 158 break; 159 } 160 if(mach[i].in[k] == 0 && mach[j].out[k] == 1){ 161 162 ok = false; 163 break; 164 } 165 } 166 if(ok == true){ 167 addEdge(j+n, i, INF); 168 } 169 } 170 } 171 172 for(int i = 2; i <= e_n; ++ i){ 173 e_bak[i] = e[i]; 174 } 175 176 printf("%d ", dinic(src, sink)); 177 178 for(int u = n+1; u <= n*2; ++ u){ 179 for(int i = g[u]; i; i = e[i].pre){ 180 int v = e[i].v; 181 if(v == 0 || v > n) 182 continue; 183 int f = e[i].f; 184 int f_bak = e_bak[i].f; 185 if(f_bak > f){ 186 f_e ++; 187 flow[f_e].u = u-n; 188 flow[f_e].v = v; 189 flow[f_e].f = f_bak - f; 190 } 191 } 192 } 193 printf("%d\n", f_e); 194 for(int i = 1; i <= f_e; ++ i){ 195 printf("%d %d %d\n", flow[i].u, flow[i].v, flow[i].f); 196 } 197 } 198 }
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原文地址:http://www.cnblogs.com/takeoffyoung/p/4658217.html