UVALive - 3353 Optimal Bus Route Design(二分图最佳匹配)

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题目大意：给出一个n个点的有向带权图，找若干个圈，使得每个节点恰好属于一个圈，要求总长度尽量小

解题思路：首先要在一个圈内，且只能在一个圈内，那么每个点的入度和出度要都为1，所以二分图就有了
接着连边，给出来的边都连了，没给的边就设为INF，最后求最小权和就可以来
如果最小权和大于INF了，就表示无法满足每个点都在一个圈内的要求了
注意：会有重边

#include <cstdio>#include <cstring>#define N 110#define INF 0x3f3f3f3f#define max(a,b)((a)>(b)?(a):(b))#define min(a,b)((a)<(b)?(a):(b))#define abs(a)((a)>0?(a):(-(a)))#define ll long longint w[N][N], Lx[N], Ly[N], left[N], slack[N];bool S[N], T[N];int n, m;void init() {    memset(w, 0x3f, sizeof(w));    int j, d;    for (int i = 1; i <= n; i++) {        while(scanf("%d", &j) && j) {            scanf("%d", &d);            w[i][j] = min(w[i][j], d);        }    }    for (int i = 1; i <= n; i++)        for (int j = 1; j <= n; j++)            w[i][j] = -w[i][j];}bool match(int i) {    S[i] = true;    for (int j = 1; j <= n; j++) {        if (Lx[i] + Ly[j] == w[i][j] && !T[j]) {            T[j] = true;            if (!left[j] || match(left[j])) {                left[j] = i;                return true;            }        }        else slack[j] = min(slack[j], Lx[i] + Ly[j] - w[i][j]);    }    return false;}int KM() {    for (int i = 1; i <= n; i++) {        left[i] = Ly[i] = 0;        Lx[i] = -INF;        for (int j = 1; j <= n; j++)            Lx[i] = max(Lx[i], w[i][j]);    }    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= n; j++)            slack[j] = INF;        while (1) {            for (int j = 1; j <= n; j++) S[j] = T[j] = 0;            if (match(i)) break;            int a = INF;            for (int j = 1; j <= n; j++)                if (!T[j])                    a = min(a, slack[j]);            for (int j = 1; j <= n; j++) {                if (S[j]) Lx[j] -= a;                if (T[j]) Ly[j] += a;            }        }    }    ll ans = 0;    for (int i = 1; i <= n; i++)        ans += Lx[i] + Ly[i];    return ans;}void solve() {    ll ans = KM();    if (abs(ans) >= INF)        printf("N\n");    else        printf("%lld\n", abs(ans));}int main() {    while (scanf("%d", &n) != EOF && n) {        init();        solve();    }    return 0;}

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