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