二分图练习小结

最新推荐文章于 2024-08-01 22:05:10 发布

转载最新推荐文章于 2024-08-01 22:05:10 发布 · 116 阅读

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原文链接：http://www.cnblogs.com/andybear/p/4405073.html

本文解析了三道图论算法题：AirRaid题求有向无环图的最小路径覆盖；Ollivanders题实现棍子和人的最大匹配；棋盘游戏题通过二分图匹配确定关键位置。

1. HDU 1151 Air Raid

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;
    }
}

3. HDU 1281 棋盘游戏

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);
    }
}

转载于:https://www.cnblogs.com/andybear/p/4405073.html