hdu6346: 整数规划

读完题目就想费用流和单纯形去了。

实际上，我们发现这个题目的求解问题。

其实就是KM算法求二分图最小权匹配的做法，跑个KM就行了。

然后这个题写递归的KM根本过不了，得用非递归的。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define inf 1000000000000ll
inline int read() {
    char ch = getchar(); int x = 0, f = 1;
    while(ch < '0' || ch > '9') {
        if(ch == '-') f = -1;
        ch = getchar();
    }
    while('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
namespace KM{
    #define M 410
    int n;
    LL A[M], B[M], mn[M];
    int w[M][M], lk[M], way[M];
    bool vis[M];
    inline void km(){
        memset(lk, -1, sizeof(lk));
        for(int x = 1; x <= n; ++ x) {
            lk[0] = x;
            int j0 = 0;
            memset(vis, 0, sizeof(vis));
            memset(mn, 0x3f, sizeof(mn));
            do{
                vis[j0] = true;
                int i0 = lk[j0], j1; 
                LL num = inf;
                for(int j = 1; j <= n; ++ j){
                    if(!vis[j]){
                        LL t = A[i0] + B[j] - w[i0][j];
                        if(t < mn[j]) mn[j] = t, way[j] = j0;
                        if(mn[j] < num) num = mn[j], j1 = j;
                    }
                }
                for(int j = 0; j <= n; ++ j){
                    if(vis[j]) A[lk[j]] -= num, B[j] += num;
                    else mn[j] -= num;
                } 
                //cerr << x <<endl;
                j0 = j1;
            } while(~lk[j0]);
            do{
                int j1 = way[j0];
                lk[j0] = lk[j1];
                j0 = j1;
            } while(j0);    
        }
    }
    inline LL getsum(){
        LL ret = 0;
        for(int i = 1; i <= n; ++ i) ret += A[i], ret += B[i];
        return ret;
    }
} 
signed main() {
    //int TT = clock();
    int T;
    scanf("%d", &T);
    for(int Ca = 1; Ca <= T; ++ Ca) {
        scanf("%d", &KM::n);
        for(int i = 1; i <= KM::n; ++ i) {
            for(int j = 1; j <= KM::n; ++ j) {
                KM::w[i][j] = read();
                KM::w[i][j] *= -1;
            }
        }
        //cerr << 2333 << endl;
        KM::km();
        printf("Case #%d: %I64d
", Ca, -KM::getsum());
    }
    //cerr << clock() - TT;
}