UVa 二分图匹配 Biginners

UVa 1045 - The Great Wall Game 最小权匹配

题意：给你一个n*n的棋盘，上面有n个棋子，要求通过移动各个棋子使得棋子在同一行或者同一列或者对角线上，求最小移动次数。

思路：直接对于所有可能情况构造二分图，X集合为最初棋子，Y集合为移动后的棋子方位，边权为移动的次数。然后KM算法求最小权匹配。

/* **********************************************
Author      : JayYe
Created Time: 2013-8-18 15:55:41
File Name   : zzz.cpp
 *********************************************** */

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int max(int a, int b) { return a>b?a:b; }
int min(int a, int b) { return a>b?b:a; }

const int maxn = 22;

struct PP {
    int x, y;
}a[maxn];

int n, slack[maxn], lx[maxn], ly[maxn], match[maxn], w[maxn][maxn];
bool S[maxn], T[maxn];

bool dfs(int i) {
    S[i] = 1;
    for(int j = 1;j <= n; j++) if(!T[j])
        slack[j] = min(slack[j], w[i][j] - lx[i] - ly[j]);
    for(int j = 1;j <= n; j++) if(w[i][j] == lx[i] + ly[j] && !T[j]) {
        T[j] = 1;
        if(!match[j] || dfs(match[j])) {
            match[j] = i;
            return true;
        }
    }
    return false;
}

void update() {
    int delta = 1<<22;
    for(int i = 1;i <= n; i++) if(!T[i])
        delta = min(delta, slack[i]);
    for(int i = 1;i <= n; i++) {
        if(S[i])    lx[i] += delta;
        if(T[i])    ly[i] -= delta;
    }
}

void KM() {
    int i, j;
    for(i = 1;i <= n; i++) {
        ly[i] = match[i] = 0;
        lx[i] = 1<<22;
        for(j = 1;j <= n; j++)
            lx[i] = min(lx[i], w[i][j]);
    }
    for(i = 1;i <= n; i++) {
        while(true) {
            for(j = 1;j <= n; j++) S[j] = T[j] = 0, slack[j] = 1<<22;
            if(dfs(i))  break;
            else    update();
        }
    }
}

int solve() {
    int i, j, k;

    for(i = 1;i <= n; i++)
        scanf("%d%d" ,&a[i].x, &a[i].y);
    int ans = 1<<22;
    // 棋子在同一行的情况
    for(i = 1;i <= n; i++) {
        for(j = 1;j <= n; j++)
            for(k = 1;k <= n; k++)
                w[j][k] = abs(i - a[j].x) + abs(k - a[j].y);
        KM();
        int cur = 0;
        for(j = 1;j <= n; j++)  cur += lx[j] + ly[j];
        ans = min(ans, cur);
    }
    // 棋子在同一列的情况
    for(i = 1;i <= n; i++) {
        for(j = 1;j <= n; j++)
            for(k = 1;k <= n; k++)
                w[j][k] = abs(k - a[j].x) + abs(i - a[j].y);
        KM();
        int cur = 0;
        for(j = 1;j <= n; j++)  cur += lx[j] + ly[j];
        ans = min(ans, cur);
    }
    // 棋子在对角线的两种情况
    for(i = 1;i <= n; i++)
        for(j = 1;j <= n; j++)
            w[i][j] = abs(j - a[i].x) + abs(j - a[i].y);
    KM();
    int cur = 0;
    for(i = 1;i <= n; i++)  cur += lx[i] + ly[i];
    ans = min(ans, cur);

    for(i = 1;i <= n; i++)
        for(j =1;j <= n; j++)
            w[i][j] = abs(j - a[i].x) + abs(n-j+1 - a[i].y);
    KM();
    cur = 0;
    for(i = 1;i <= n; i++)  cur += lx[i] + ly[i];
    ans = min(ans, cur);
    return ans;
}

int main() {
    int cas = 1;
    while(scanf("%d", &n) != -1  && n) {
        printf("Board %d: %d moves required.

", cas++, solve());
    }
    return 0;
}

UVa 12168 - Cat vs. Dog 最大独立集

根据题意直接构造二分图，X集合表示喜欢狗的人，Y集合表示喜欢猫的人，如果X集合里的人不喜欢某只猫，就与Y集合里喜欢该猫的人连边，反之也一样。那么要使得尽可能多的人满足愿望，也就是求最大独立集。

最大独立集 = N - 最大匹配

/* **********************************************
Author      : JayYe
Created Time: 2013-8-18 16:53:58
File Name   : zzz.cpp
*********************************************** */

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int maxn = 500+10;
struct PP {
    int x, y;
    PP() {}
    PP(int x, int y) : x(x), y(y) {}
}cat[maxn], dog[maxn];

int n, m, match[maxn];
bool vis[maxn], mp[maxn][maxn];

bool dfs(int i) {
    for(int j = 1;j <= m; j++) if(mp[i][j] && !vis[j]) {
        vis[j] = 1;
        if(!match[j] || dfs(match[j])) {
            match[j] = i;
            return true;
        }
    }
    return false;
}

int main() {
    int i, j, t, c, d;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d%d", &c, &d, &n);
        int n1 = 0, n2 = 0, x, y;
        for(i = 1;i <= n; i++) {
            char ch1[11], ch2[11];
            scanf("%s%s", ch1, ch2);
            if(ch1[0] == 'C') {
                sscanf(ch1+1, "%d", &x);
                sscanf(ch2+1, "%d", &y);
                cat[++n1] = PP(x, y);
            }
            else {
                sscanf(ch1+1, "%d", &x);
                sscanf(ch2+1, "%d", &y);
                dog[++n2] = PP(x, y);
            }
        }
        n = n1, m = n2;
        if(n == 0 || m == 0) {
            printf("%d
", n+m); continue;
        }
        for(i = 1;i <= n; i++) {
            for(j = 1;j <= m; j++) {
                if(cat[i].y == dog[j].x || cat[i].x == dog[j].y)
                    mp[i][j] = 1;
                else
                    mp[i][j] = 0;
            }
        }
        for(i = 1;i <= m; i++)  match[i] = 0;
        int ans = 0;
        for(i = 1;i <= n; i++) {
            for(j = 1;j <= m; j++)  vis[j] = 0;
            if(dfs(i))  ans++;
        }
        printf("%d
", n+m - ans);
    }
    return 0;
}

Uva 1349 - Optimal Bus Route Design

题意：给你一个n个点的有向带环图，要你找出几个圈使得每个结点只属于一个圈，要求输出最小的总的长度，如果没有这样的方案，输出N。

构造二分图，把所有的结点拆成两个，放在X集合的为i， 放在Y集合的为 i '，如果有边i -> j，则在图中引入边i -> j'，这样子构造好后实际上就是求最小权完美匹配，如果没有完美匹配则无解。

这里需要注意的是输入的边可能有好多条是重复的但是权值不同，需要取最小权，这个wa了我好几发。。。

/* **********************************************
Author      : JayYe
Created Time: 2013-8-18 17:31:11
File Name   : zzz.cpp
 *********************************************** */

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int max(int a, int b) { return a>b?a:b; }
int min(int a, int b) { return a>b?b:a; }

const int INF = 1<<22;
const int maxn = 100+10;
int n, slack[maxn], match[maxn], lx[maxn], ly[maxn], w[maxn][maxn];
bool S[maxn], T[maxn];

bool dfs(int i) {
    S[i] = 1;
    for(int j = 1;j <= n; j++) if(!T[j])
        slack[j] = min(slack[j], w[i][j] - lx[i] - ly[j]);
    for(int j = 1;j <= n; j++) if(w[i][j] == lx[i] + ly[j] && !T[j]) {
        T[j] = 1;
        if(!match[j] || dfs(match[j])) {
            match[j] = i;
            return true;
        }
    }
    return false;
}

void update() {
    int delta = INF;
    for(int i = 1;i <= n; i++) if(!T[i])
        delta = min(delta, slack[i]);
    for(int i = 1;i <= n; i++) {
        if(S[i])    lx[i] += delta;
        if(T[i])    ly[i] -= delta;
    }
}

void KM() {
    int i, j;
    for(i = 1;i <= n; i++) {
        ly[i] = match[i] = 0;
        lx[i] = INF;
        for(j = 1;j <= n; j++)
            lx[i] = min(lx[i], w[i][j]);
    }

    for(i = 1;i <= n; i++) {
        while(true) {
            for(j = 1;j <= n; j++)  S[j] = T[j] = 0, slack[j] = INF;
            if(dfs(i))  break;
            else    update();
        }
    }
}

void solve() {
    int i, j;
    for(i = 1;i <= n; i++)
        for(j = 1;j <= n; j++)
            w[i][j] = INF;
    for(i = 1;i <= n; i++) {
        while(scanf("%d", &j) && j) {
            int dis;
            scanf("%d", &dis);
            // 同一条路要取最小值
            w[i][j] = min(w[i][j], dis);
        }
    }
    KM();
    int ans = 0;
    for(i = 1;i <= n; i++)  ans += lx[i] + ly[i];
    if(ans > INF-10)   puts("N");
    else    printf("%d
", ans);
}

int main() {
    while(scanf("%d", &n) != -1 && n) {
        solve();
    }
    return 0;
}