飞行员配对方案问题 网络流24题
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输入输出样例
输入 #1
5 10 1 7 1 8 2 6 2 9 2 10 3 7 3 8 4 7 4 8 5 10 -1 -1
输出 #1
View Code
4 1 7 2 9 3 8 5 10
思路
题目给出两个不同阵营
不同阵营之间的人可以组队
一个人只能和一个人组队
求最大匹配数
显然可以直接偷懒上匈牙利
但既然叫网络流24题就用网络流做吧
建立一个S、一个T表示超级源、汇
把A阵营的人连S
B阵营的人连T
认识的人之间就可以连边
因为每个人都只能和一个另外阵营的人组队
说明每个点都只能最多用一次
所以这些边的 capacity 都是 1
图建完后,如何输出他们匹配的方案呢
考虑最小路径那样记录
但是由于悔边的原因并不好记录
所以直接从悔边下手
当一对组合成立
说明正向边容量 - 1
则悔边容量应该 + 1
所以只要是悔边容量变大了的
那么连着这条边的两个人都是被选中的
CODE
1 #include <bits/stdc++.h> 2 #define dbg(x) cout << #x << "=" << x << endl 3 #define eps 1e-8 4 #define pi acos(-1.0) 5 6 using namespace std; 7 typedef long long LL; 8 9 const int inf = 0x3f3f3f3f; 10 11 template<class T>inline void read(T &res) 12 { 13 char c;T flag=1; 14 while((c=getchar())<‘0‘||c>‘9‘)if(c==‘-‘)flag=-1;res=c-‘0‘; 15 while((c=getchar())>=‘0‘&&c<=‘9‘)res=res*10+c-‘0‘;res*=flag; 16 } 17 18 namespace _buff { 19 const size_t BUFF = 1 << 19; 20 char ibuf[BUFF], *ib = ibuf, *ie = ibuf; 21 char getc() { 22 if (ib == ie) { 23 ib = ibuf; 24 ie = ibuf + fread(ibuf, 1, BUFF, stdin); 25 } 26 return ib == ie ? -1 : *ib++; 27 } 28 } 29 30 int qread() { 31 using namespace _buff; 32 int ret = 0; 33 bool pos = true; 34 char c = getc(); 35 for (; (c < ‘0‘ || c > ‘9‘) && c != ‘-‘; c = getc()) { 36 assert(~c); 37 } 38 if (c == ‘-‘) { 39 pos = false; 40 c = getc(); 41 } 42 for (; c >= ‘0‘ && c <= ‘9‘; c = getc()) { 43 ret = (ret << 3) + (ret << 1) + (c ^ 48); 44 } 45 return pos ? ret : -ret; 46 } 47 48 const int maxn = 200007; 49 50 int n, m; 51 int s, t; 52 53 struct edge{ 54 int from,to; 55 LL cap,flow; 56 }; 57 58 map<int, int> vis; 59 60 struct DINIC { 61 int head[maxn << 1], nxt[maxn << 1], edge[maxn << 1], cnt; 62 int cap[maxn << 1], depth[maxn << 1]; 63 64 void init() { 65 cnt = 1; 66 memset(head, 0, sizeof(head)); 67 } 68 69 void BuildGraph(int u, int v, int w) { 70 ++cnt; 71 edge[cnt] = v; 72 nxt[cnt] = head[u]; 73 cap[cnt] = w; 74 head[u] = cnt; 75 76 ++cnt; 77 edge[cnt] = u; 78 nxt[cnt] = head[v]; 79 cap[cnt] = 0; 80 head[v] = cnt; 81 } 82 83 queue<int> q; 84 85 bool bfs() { 86 memset(depth, 0, sizeof(depth)); 87 depth[s] = 1; 88 q.push(s); 89 while(!q.empty()) { 90 int u = q.front(); 91 q.pop(); 92 for ( int i = head[u]; i; i = nxt[i] ) { 93 int v = edge[i]; 94 if(depth[v]) { 95 continue; 96 } 97 if(cap[i]) { 98 depth[v] = depth[u] + 1; 99 q.push(v); 100 } 101 } 102 } 103 return depth[t]; 104 } 105 106 int dfs(int u, int dist) { 107 if(u == t) { 108 return dist; 109 } 110 int flow = 0; 111 for ( int i = head[u]; i && dist; i = nxt[i] ) { 112 if(cap[i] == 0) 113 continue; 114 int v = edge[i]; 115 if(depth[v] != depth[u] + 1) { 116 continue; 117 } 118 int res = dfs(v, min(cap[i], dist)); 119 cap[i] -= res; 120 cap[i ^ 1] += res; 121 //printf("cap[%d]:%d ",t, cap[t]); 122 vis[u] = v; 123 dist -= res; 124 flow += res; 125 } 126 return flow; 127 } 128 129 int maxflow() { 130 int ans = 0; 131 while(bfs()) { 132 ans += dfs(s, inf); 133 } 134 return ans; 135 } 136 } dinic; 137 138 int main() 139 { 140 //freopen("data.txt", "r", stdin); 141 read(m); read(n); 142 int u, v; 143 s = 0, t = n + m + 1; 144 dinic.init(); 145 while(1) { 146 read(u); read(v); 147 if(u == -1 && v == -1) { 148 break; 149 } 150 dinic.BuildGraph(u, v, 1); 151 } 152 for ( int i = 1; i <= m; ++i ) { 153 dinic.BuildGraph(s, i, 1); 154 } 155 for ( int i = m + 1; i <= m + n; ++i ) { 156 dinic.BuildGraph(i, t, 1); 157 } 158 cout << dinic.maxflow() << endl; 159 int tot = dinic.cnt; 160 for ( int i = 2; i <= tot; i += 2 ) { 161 int u = dinic.edge[i]; 162 int v = dinic.edge[i ^ 1]; 163 if(u == s || u == t || v == s || v == t) { 164 continue; 165 } 166 else { 167 int cap = dinic.cap[i ^ 1]; 168 if(cap != 0) { 169 printf("%d %d ",v, u); 170 } 171 } 172 } 173 return 0; 174 }
#include <bits/stdc++.h>
#define dbg(x) cout << #x << "=" << x << endl
#define eps 1e-8
#define pi acos(-1.0)
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
template<class T>inline void read(T &res)
{
char c;T flag=1;
while((c=getchar())<‘0‘||c>‘9‘)if(c==‘-‘)flag=-1;res=c-‘0‘;
while((c=getchar())>=‘0‘&&c<=‘9‘)res=res*10+c-‘0‘;res*=flag;
}
namespace _buff {
const size_t BUFF = 1 << 19;
char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
char getc() {
if (ib == ie) {
ib = ibuf;
ie = ibuf + fread(ibuf, 1, BUFF, stdin);
}
return ib == ie ? -1 : *ib++;
}
}
int qread() {
using namespace _buff;
int ret = 0;
bool pos = true;
char c = getc();
for (; (c < ‘0‘ || c > ‘9‘) && c != ‘-‘; c = getc()) {
assert(~c);
}
if (c == ‘-‘) {
pos = false;
c = getc();
}
for (; c >= ‘0‘ && c <= ‘9‘; c = getc()) {
ret = (ret << 3) + (ret << 1) + (c ^ 48);
}
return pos ? ret : -ret;
}
const int maxn = 200007;
int n, m;
int s, t;
struct edge{
int from,to;
LL cap,flow;
};
map<int, int> vis;
struct DINIC {
int head[maxn << 1], nxt[maxn << 1], edge[maxn << 1], cnt;
int cap[maxn << 1], depth[maxn << 1];
void init() {
cnt = 1;
memset(head, 0, sizeof(head));
}
void BuildGraph(int u, int v, int w) {
++cnt;
edge[cnt] = v;
nxt[cnt] = head[u];
cap[cnt] = w;
head[u] = cnt;
++cnt;
edge[cnt] = u;
nxt[cnt] = head[v];
cap[cnt] = 0;
head[v] = cnt;
}
queue<int> q;
bool bfs() {
memset(depth, 0, sizeof(depth));
depth[s] = 1;
q.push(s);
while(!q.empty()) {
int u = q.front();
q.pop();
for ( int i = head[u]; i; i = nxt[i] ) {
int v = edge[i];
if(depth[v]) {
continue;
}
if(cap[i]) {
depth[v] = depth[u] + 1;
q.push(v);
}
}
}
return depth[t];
}
int dfs(int u, int dist) {
if(u == t) {
return dist;
}
int flow = 0;
for ( int i = head[u]; i && dist; i = nxt[i] ) {
if(cap[i] == 0)
continue;
int v = edge[i];
if(depth[v] != depth[u] + 1) {
continue;
}
int res = dfs(v, min(cap[i], dist));
cap[i] -= res;
cap[i ^ 1] += res;
//printf("cap[%d]:%d
",t, cap[t]);
vis[u] = v;
dist -= res;
flow += res;
}
return flow;
}
int maxflow() {
int ans = 0;
while(bfs()) {
ans += dfs(s, inf);
}
return ans;
}
} dinic;
int main()
{
//freopen("data.txt", "r", stdin);
read(m); read(n);
int u, v;
s = 0, t = n + m + 1;
dinic.init();
while(1) {
read(u); read(v);
if(u == -1 && v == -1) {
break;
}
dinic.BuildGraph(u, v, 1);
}
for ( int i = 1; i <= m; ++i ) {
dinic.BuildGraph(s, i, 1);
}
for ( int i = m + 1; i <= m + n; ++i ) {
dinic.BuildGraph(i, t, 1);
}
cout << dinic.maxflow() << endl;
int tot = dinic.cnt;
for ( int i = 2; i <= tot; i += 2 ) {
int u = dinic.edge[i];
int v = dinic.edge[i ^ 1];
if(u == s || u == t || v == s || v == t) {
continue;
}
else {
int cap = dinic.cap[i ^ 1];
if(cap != 0) {
printf("%d %d
",v, u);
}
}
}
return 0;
}
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