【HDOJ】2853 Assignment

最小费用最大流可解最优解。至于dif如何解，可以把w扩大100倍，如果mission编号和排列P相等则对w+1，然后建立网络流。
对结果取模100可以得到没有改变mission的company数目，用company数目减之可以得到dif.

 /* 2853 */
 #include <iostream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 const int INF = 0x1f1f1f1f;
 const int maxn = ;
 const int maxv = maxn * ;
 const int maxe = maxv * maxv * ;
 int V[maxe], F[maxe], W[maxe], nxt[maxe];
 int head[maxv], dis[maxv], pre[maxv], ID[maxv];
 bool visit[maxv];
 int M[maxn][maxn], P[maxn];
 int s, t, m;

 void addEdge(int u, int v, int f, int w) {
     V[m] = v;
     F[m] = f;
     W[m] = w;
     nxt[m] = head[u];
     head[u] = m++;

     V[m] = u;
     F[m] = ;
     W[m] = -w;
     nxt[m] = head[v];
     head[v] = m++;
 }

 bool bfs() {
     queue<int> Q;
     int u, v, k;

     memset(dis, INF, sizeof(dis));
     memset(visit, false, sizeof(visit));
     Q.push(s);
     dis[s] = ;

     while (!Q.empty()) {
         u = Q.front();
         Q.pop();
         visit[u] = false;
         for (k=head[u]; k!=-; k=nxt[k]) {
             v = V[k];
             if (F[k] && dis[v]>dis[u]+W[k]) {
                 dis[v] = dis[u] + W[k];
                 ID[v] = k;
                 pre[v] = u;
                 if (!visit[v]) {
                     visit[v] = true;
                     Q.push(v);
                 }
             }
         }
     }

     return dis[t]==INF;
 }

 int MCMF() {
     int ret = , tmp;
     int u, v, k;

     while () {
         if (bfs())
             break;

         tmp = INF;
         for (v=t, u=pre[v]; v!=s; v=u, u=pre[v]) {
             k = ID[v];
             tmp = min(F[k], tmp);
         }

         for (v=t, u=pre[v]; v!=s; v=u, u=pre[v]) {
             k = ID[v];
             F[k] -= tmp;
             F[k^] += tmp;
         }

         ret += dis[t] * tmp;
     }

     return ret;
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     int r, c;
     int ans, tot, tmp, dif;

     while (scanf("%d %d", &r, &c) != EOF) {

         s = m = ;
         t = r + c + ;
         memset(head, -, sizeof(head));

         rep(i, , r+)
             rep(j, , c+)
                 scanf("%d", &M[i][j]);
         rep(i, , r+)
             scanf("%d", &P[i]);

         rep(i, , r+)
             addEdge(s, i, , );

         rep(j, , c+)
             addEdge(j+r, t, , );

         tot = ;
         rep(i, , r+) {
             rep(j, , c+) {
                 if (j == P[i]) {
                     tmp = M[i][j] *  + ;
                 } else {
                     tmp = M[i][j] * ;
                 }
                 addEdge(i, j+r, , -tmp);
             }
             tot += M[i][P[i]];
         }

         tmp = -MCMF();
         ans = tmp/ - tot;
         #ifndef ONLINE_JUDGE
             printf("tot = %d, MCMF = %d\n", tot, tmp);
         #endif
         dif = r - tmp%;
         printf("%d %d\n", dif, ans);
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }