标签：inpu php rem ESS bool some enc person bsp
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9289 Accepted Submission(s): 3019
1 /* 2 结点0 ~ n - 1存左牛结点 3 结点n ~ 2 * n - 1存右牛结点 4 结点2 * n ~ 2 * n + f - 1存左食物 5 结点2 * n + f ~ 2 * n + f * 2 - 1存右食物 6 结点2 * n + 2 * f ~ 2 * n + 2 * f + d - 1存饮料左 7 结点2 * n + 2 * f + d ~ 2 * n + 2 * f + d * 2 - 1存饮料右 8 结点2 * n + 2 * f + 2 * d为s，t = s = 1。 9 */ 10 11 #include <cstdio> 12 #include <cstring> 13 #include <algorithm> 14 using namespace std; 15 16 const int maxn = 200 + 5, maxm = 1000 * 1000 + 5, inf = 0x3f3f3f3f; 17 int numf[maxn], numd[maxn]; 18 char str[maxn]; 19 20 int tot, head[maxn << 3], que[maxn << 3], dep[maxn << 3], cur[maxn << 3], sta[maxn << 3]; 21 22 struct Edge { 23 int to, cap, flow, next, from; 24 } edge[maxm << 1]; 25 26 void init() { 27 tot = 2; 28 memset(head, -1, sizeof head); 29 } 30 31 void addedge(int u, int v, int w) { 32 edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0; edge[tot].from = u; 33 edge[tot].next = head[u]; head[u] = tot ++; 34 edge[tot].to = u; edge[tot].cap = 0; edge[tot].flow = 0; edge[tot].from = v; 35 edge[tot].next = head[v]; head[v] = tot ++; 36 } 37 38 bool bfs(int s, int t, int n) { 39 memset(dep, -1, sizeof dep[0] * (n + 1)); 40 int front = 0, tail = 0; 41 dep[s] = 0; 42 que[tail ++] = s; 43 while(front < tail) { 44 int u = que[front ++]; 45 for(int i = head[u]; ~i; i = edge[i].next) { 46 int v = edge[i].to; 47 if(edge[i].cap > edge[i].flow && dep[v] == -1) { 48 dep[v] = dep[u] + 1; 49 if(v == t) return true; 50 que[tail ++] = v; 51 } 52 } 53 } 54 return false; 55 } 56 57 int dinic(int s, int t, int n) { 58 int maxflow = 0; 59 while(bfs(s, t, n)) { 60 for(int i = 0; i <= n; i ++) cur[i] = head[i]; 61 int u = s, tail = 0; 62 while(~cur[s]) { 63 if(u == t) { 64 int tp = inf; 65 for(int i = tail - 1; i >= 0; i --) 66 tp = min(tp, edge[sta[i]].cap - edge[sta[i]].flow); 67 maxflow += tp; 68 for(int i = tail - 1; i >= 0; i --) { 69 edge[sta[i]].flow += tp; 70 edge[sta[i] ^ 1].flow -= tp; 71 if(edge[sta[i]].cap - edge[sta[i]].flow == 0) tail = i; 72 } 73 u = edge[sta[tail] ^ 1].to; 74 } else if(~ cur[u] && edge[cur[u]].cap > edge[cur[u]].flow && dep[u] + 1 == dep[edge[cur[u]].to]) { 75 sta[tail ++] = cur[u]; 76 u = edge[cur[u]].to; 77 } else { 78 while(u != s && cur[u] == -1) 79 u = edge[sta[-- tail] ^ 1].to; 80 cur[u] = edge[cur[u]].next; 81 } 82 } 83 } 84 return maxflow; 85 } 86 87 int main() { 88 int n, f, d; 89 while(~scanf("%d %d %d", &n, &f, &d)) { 90 init(); 91 int s = 2 * n + 2 * f + d * 2, t = s + 1;//超级源点和超级汇点 92 for(int i = 0; i < f; i ++) { 93 scanf("%d", &numf[i]); 94 addedge(s, 2 * n + i, inf);//超级源点到食物左 95 addedge(2 * n + i, 2 * n + f + i, numf[i]);//食物左到食物右 96 } 97 for(int i = 0; i < d; i ++) { 98 scanf("%d", &numd[i]); 99 addedge(2 * n + 2 * f + i, 2 * n + 2 * f + d + i, numd[i]);//饮料左到饮料右 100 addedge(2 * n + 2 * f + d + i, t, inf);//饮料右到超级汇点 101 } 102 for(int i = 0; i < n; i ++) { 103 addedge(i, n + i, 1);//左牛到右牛 104 } 105 for(int i = 0; i < n; i++) { 106 scanf("%s", str); 107 for(int j = 0; j < f; j ++) { 108 if(str[j] == ‘Y‘) 109 addedge(2 * n + f + j, i, 1);//从食物右到左牛 110 } 111 } 112 for(int i = 0; i < n; i ++) { 113 scanf("%s", str); 114 for(int j = 0; j < d; j ++) { 115 if(str[j] == ‘Y‘) 116 addedge(n + i, 2 * n + 2 * f + j, 1);//从右牛到左饮料 117 } 118 } 119 // for(int i = 2; i <= tot; i ++) { 120 // printf("%d -> %d\n", edge[i].from, edge[i].to); 121 // } 122 printf("%d\n", dinic(s, t, 2 * n + 2 * f + 2 * d + 2)); 123 } 124 return 0; 125 }
下面这种建图方法和上面的类似，只是上图是拆点限制点流量，而我们知道对于每一件食物，如果我们有一个人选取它，那么它必定是只选取了一件，因为后面拆点n决定的，那么也就是每个人只能取他所喜欢食物中的一种中的一个，所以我们只需要对我们能够提供的某种食物限量就可以了，也就是从源点到某种食物权值为food_num，这样就可以限制住每种食物的用量了，接着看饮料，如果某个人选取了一个饮料，那么他也只能选取这一种饮料中的一瓶，因为前面已经对n拆点导致它能扩展的流也只有1，所以导致她选的饮料也是1对1，所以想要限制饮料的个数，也只需要无限索取，直到最后无法流到汇点就ok，那也就是从饮料到汇点权值为drink_num。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 6 const int maxn = 200 + 5, maxm = 1000 * 1000 + 5, inf = 0x3f3f3f3f; 7 int numf[maxn], numd[maxn]; 8 char str[maxn]; 9 10 int tot, head[maxn << 3], que[maxn << 3], dep[maxn << 3], cur[maxn << 3], sta[maxn << 3]; 11 12 struct Edge { 13 int to, cap, flow, next, from; 14 } edge[maxm << 1]; 15 16 void init() { 17 tot = 2; 18 memset(head, -1, sizeof head); 19 } 20 21 void addedge(int u, int v, int w) { 22 edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0; edge[tot].from = u; 23 edge[tot].next = head[u]; head[u] = tot ++; 24 edge[tot].to = u; edge[tot].cap = 0; edge[tot].flow = 0; edge[tot].from = v; 25 edge[tot].next = head[v]; head[v] = tot ++; 26 } 27 28 bool bfs(int s, int t, int n) { 29 memset(dep, -1, sizeof dep[0] * (n + 1)); 30 int front = 0, tail = 0; 31 dep[s] = 0; 32 que[tail ++] = s; 33 while(front < tail) { 34 int u = que[front ++]; 35 for(int i = head[u]; ~i; i = edge[i].next) { 36 int v = edge[i].to; 37 if(edge[i].cap > edge[i].flow && dep[v] == -1) { 38 dep[v] = dep[u] + 1; 39 if(v == t) return true; 40 que[tail ++] = v; 41 } 42 } 43 } 44 return false; 45 } 46 47 int dinic(int s, int t, int n) { 48 int maxflow = 0; 49 while(bfs(s, t, n)) { 50 for(int i = 0; i <= n; i ++) cur[i] = head[i]; 51 int u = s, tail = 0; 52 while(~cur[s]) { 53 if(u == t) { 54 int tp = inf; 55 for(int i = tail - 1; i >= 0; i --) 56 tp = min(tp, edge[sta[i]].cap - edge[sta[i]].flow); 57 maxflow += tp; 58 for(int i = tail - 1; i >= 0; i --) { 59 edge[sta[i]].flow += tp; 60 edge[sta[i] ^ 1].flow -= tp; 61 if(edge[sta[i]].cap - edge[sta[i]].flow == 0) tail = i; 62 } 63 u = edge[sta[tail] ^ 1].to; 64 } else if(~ cur[u] && edge[cur[u]].cap > edge[cur[u]].flow && dep[u] + 1 == dep[edge[cur[u]].to]) { 65 sta[tail ++] = cur[u]; 66 u = edge[cur[u]].to; 67 } else { 68 while(u != s && cur[u] == -1) 69 u = edge[sta[-- tail] ^ 1].to; 70 cur[u] = edge[cur[u]].next; 71 } 72 } 73 } 74 return maxflow; 75 } 76 77 int main() { 78 int n, f, d; 79 while(~scanf("%d %d %d", &n, &f, &d)) { 80 init(); 81 int s = 2 * n + f + d, t = s + 1;//超级源点和超级汇点 82 for(int i = 0; i < f; i ++) { 83 scanf("%d", &numf[i]); 84 addedge(s, n * 2 + i, numf[i]); 85 } 86 for(int i = 0; i < d; i ++) { 87 scanf("%d", &numd[i]); 88 addedge(n * 2 + f + i, t, numd[i]); 89 } 90 for(int i = 0; i < n; i++) { 91 addedge(i, n + i, 1);//左牛到右牛 92 scanf("%s", str); 93 for(int j = 0; j < f; j ++) { 94 if(str[j] == ‘Y‘) 95 addedge(2 * n + j, i, 1); 96 } 97 } 98 for(int i = 0; i < n; i ++) { 99 scanf("%s", str); 100 for(int j = 0; j < d; j ++) { 101 if(str[j] == ‘Y‘) 102 addedge(n + i, 2 * n + f + j, 1);//从右牛到饮料 103 } 104 } 105 // for(int i = 2; i <= tot; i ++) { 106 // printf("%d -> %d\n", edge[i].from, edge[i].to); 107 // } 108 printf("%d\n", dinic(s, t, 2 * n + f + d + 2)); 109 } 110 return 0; 111 }
不得不承认这种做法确实节省了很多空间呀。
标签：inpu php rem ESS bool some enc person bsp
原文地址：https://www.cnblogs.com/bianjunting/p/11403334.html
