题目描述
问题描述：
假设一个试题库中有 nn 道试题。每道试题都标明了所属类别。同一道题可能有多个类别属性。现要从题库中抽取 mm 道题组成试卷。并要求试卷包含指定类型的试题。试设计一个满足要求的组卷算法。
编程任务：
对于给定的组卷要求，计算满足要求的组卷方案。
输入格式
第一行有两个正整数 kk 和 nn。kk 表示题库中试题类型总数，nn 表示题库中试题总数。
第二行有 kk 个正整数，第 ii 个正整数表示要选出的类型 ii 的题数。这 kk 个数相加就是要选出的总题数 mm。
接下来的 nn 行给出了题库中每个试题的类型信息。每行的第一个正整数 pp 表明该题可以属于 pp 类，接着的 pp 个数是该题所属的类型号。
输出格式
输出共 kk 行，第 ii 行输出 i: 后接类型 ii 的题号。
如果有多个满足要求的方案，只要输出一个方案。
如果问题无解，则输出No Solution!。
输入输出样例
输入 #1
3 15 3 3 4 2 1 2 1 3 1 3 1 3 1 3 3 1 2 3 2 2 3 2 1 3 1 2 1 2 2 1 2 2 1 3 2 1 2 1 1 3 1 2 3
输出 #1
View Code
1: 1 6 8 2: 7 9 10 3: 2 3 4 5
思路
对于这种有限制的分配问题很容易想到用网络流建图解决
对于这道题，显然可以把试题和类型看做成二部图的两个部分
因为每道题只可以选一次
显然这张图的边容量最大为1
然后对整张图跑最大流
如果最大流 == 需要的题数
说明有方案
如何输出方案
类似于前面那道飞行员问题
因为容量为1，只要这条边被利用过，
悔边的容量一定为1.
所以只需要枚举所有被用到的类型的悔边即可
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 dist -= res; 123 flow += res; 124 } 125 return flow; 126 } 127 128 int maxflow() { 129 int ans = 0; 130 while(bfs()) { 131 ans += dfs(s, inf); 132 } 133 return ans; 134 } 135 } dinic; 136 137 int main() 138 { 139 //freopen("data.txt", "r", stdin); 140 int k; 141 read(k); read(n); 142 dinic.init(); 143 s = 0, t = n + k + 1; 144 int x; 145 for ( int i = 1; i <= k; ++i ) { 146 read(x); 147 m += x; 148 dinic.BuildGraph(n + i, t, x); 149 } 150 for ( int i = 1; i <= n; ++i ) { 151 dinic.BuildGraph(s, i, 1); 152 int p; 153 read(p); 154 for ( int j = 1; j <= p; ++j ) { 155 read(x); 156 dinic.BuildGraph(i, x + n, 1); 157 } 158 } 159 int tot = dinic.maxflow(); 160 //dbg(tot); 161 if(tot != m) { 162 puts("No Solution!"); 163 } 164 else { 165 for ( int i = 1; i <= k; ++i ) { 166 int u = n + i; 167 printf("%d:",i); 168 for ( int j = dinic.head[u]; j; j = dinic.nxt[j] ) { 169 int v = dinic.edge[j]; 170 if(dinic.cap[j] && v <= n) { 171 printf(" %d",v); 172 } 173 } 174 printf(" "); 175 } 176 } 177 return 0; 178 }
#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]);
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);
int k;
read(k); read(n);
dinic.init();
s = 0, t = n + k + 1;
int x;
for ( int i = 1; i <= k; ++i ) {
read(x);
m += x;
dinic.BuildGraph(n + i, t, x);
}
for ( int i = 1; i <= n; ++i ) {
dinic.BuildGraph(s, i, 1);
int p;
read(p);
for ( int j = 1; j <= p; ++j ) {
read(x);
dinic.BuildGraph(i, x + n, 1);
}
}
int tot = dinic.maxflow();
//dbg(tot);
if(tot != m) {
puts("No Solution!");
}
else {
for ( int i = 1; i <= k; ++i ) {
int u = n + i;
printf("%d:",i);
for ( int j = dinic.head[u]; j; j = dinic.nxt[j] ) {
int v = dinic.edge[j];
if(dinic.cap[j] && v <= n) {
printf(" %d",v);
}
}
printf("
");
}
}
return 0;
}