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题目描述有点长，就是说一个有向无环图，最少需要走几次能把整张图上的边遍历。典型的最小路径覆盖，答案就是顶点数-最大匹配数
#include <set>
#include <map>
#include <list>
#include <stack>
#include <queue>
#include <ctime>
#include <cmath>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define IT iterator
#define PB(x) push_back(x)
#define CLR(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<ull> vull;
typedef set<int> sint;
typedef set<ull> sull;
const int maxn = 150;
vint g[maxn];
int match[maxn];
bool vis[maxn];
int n,m;
bool findpath(int u) {
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (!vis[v]) {
vis[v] = 1;
if (match[v] == -1 || findpath(match[v])) {
match[v] = u;
return true;
}
}
}
return false;
}
int hungary() {
int ans = 0;
CLR(match,-1);
for (int i = 1; i <= n; i++) {
CLR(vis,0);
if (findpath(i)) ans++;
}
return ans;
}
int main() {
int T;
cin>>T;
while (T--) {
CLR(vis,0);
for (int i = 0; i < maxn; i++) g[i].clear();
cin>>n>>m;
for (int i = 0; i < m; i++) {
int a,b;
scanf("%d%d",&a,&b);
g[a].PB(b);
}
cout<<n - hungary()<<endl;
}
}
2. HDU 1179 Ollivanders: Makers of Fine Wands since 382 BC.
题目描述很长，没有看完，大致看输入输出，意思应该是棍子和人的最大匹配。
#include <set>
#include <map>
#include <list>
#include <stack>
#include <queue>
#include <ctime>
#include <cmath>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define IT iterator
#define PB(x) push_back(x)
#define CLR(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<ull> vull;
typedef set<int> sint;
typedef set<ull> sull;
const int maxn = 100 + 5;
int mk[maxn];
int nx,ny;
int cx[maxn],cy[maxn];
int g[maxn][maxn];
int path(int u) {
for (int v = 1; v <= ny; v++) {
if (g[u][v] && !mk[v]) {
mk[v] = 1;
if (cy[v] == -1 || path(cy[v])) {
cx[u] = v;
cy[v] = u;
return 1;
}
}
}
return 0;
}
int maxmatch() {
int res = 0;
CLR(cx,-1);
CLR(cy,-1);
for (int i = 1; i <= nx; i++) {
if (cx[i] == -1) {
CLR(mk,0);
res += path(i);
}
}
return res;
}
int main() {
while (cin>>nx>>ny) {
CLR(g,0);
for (int i = 1; i <= ny; i++) {
int k;
scanf("%d",&k);
for (int j = 0; j < k; j++) {
int b;
scanf("%d",&b);
g[i][b] = 1;
}
}
cout<<maxmatch()<<endl;
}
}
这个题目可以看得出来是二分图匹配，那么二分图匹配。但是建边的时候需要考虑一下，车只要是在同一行或者同一列就会相互攻击，所以可以在X坐标和Y坐标之间建边，这样的话，一个坐标对应的一行和一列上只会有一个车。解决了这个问题之后，剩下的就是模板处理计算，然后枚举依次去掉每一个点，看看最大匹配是否变化就行了。
#include <set>
#include <map>
#include <list>
#include <stack>
#include <queue>
#include <ctime>
#include <cmath>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define IT iterator
#define PB(x) push_back(x)
#define CLR(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<ull> vull;
typedef set<int> sint;
typedef set<ull> sull;
const int maxn = 100 + 5;
const int maxm = 10000 + 5;
int mk[maxn];
int nx,ny,m;
int cx[maxn],cy[maxn];
int g[maxn][maxn];
pair<int,int> P[maxm];
int path(int u) {
for (int v = 1; v <= ny; v++) {
if (g[u][v] && !mk[v]) {
mk[v] = 1;
if (cy[v] == -1 || path(cy[v])) {
cx[u] = v;
cy[v] = u;
return 1;
}
}
}
return 0;
}
int maxmatch() {
int res = 0;
CLR(cx,-1);
CLR(cy,-1);
for (int i = 1; i <= nx; i++) {
if (cx[i] == -1) {
CLR(mk,0);
res += path(i);
}
}
return res;
}
int main() {
int t = 1;
while (cin>>nx>>ny>>m) {
CLR(g,0);
for (int i = 0; i < m; i++) {
int a,b;
scanf("%d%d",&a,&b);
P[i].first = a;
P[i].second = b;
g[a][b] = 1;
}
int ans = maxmatch();
int cnt = 0;
for (int i = 0; i < m; i++) {
int a = P[i].first;
int b = P[i].second;
g[a][b] = 0;
int tmp = maxmatch();
if (tmp < ans) cnt++;
g[a][b] = 1;
}
printf("Board %d have %d important blanks for %d chessmen.\n",t++,cnt,ans);
}
}
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原文地址：http://www.cnblogs.com/andybear/p/4405073.html
