题意：一个N*N的矩阵，第i行第j列的元素大小为w[i][j]，每行求一个数row[i]，每列求一个数col[j]，使得row[i] + col[j] >= w[i][j]，且所有的row[]与所有的col[]和总和最小( N <= 500, 其它输入数为正整数且 <= 100)。
题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2378
——>>row[i] + col[j] >= w[i][j]，这个恰恰是二分图最佳完美匹配的一个式子，所以，以行row为X结点，以列col为Y结点，权值即为对应元素w[i][j]的值建图，跑一次KM就好。
另外发现：用scanf("%d", &N) == 1比用~scanf("%d", &N)快了3ms。。。
#include <cstdio>
#include <algorithm>using namespace std;const int maxn = 500 + 10;
const int INF = 0x3f3f3f3f;int N, w[maxn][maxn], lx[maxn], ly[maxn], fa[maxn];
bool S[maxn], T[maxn];bool match(int i){S[i] = 1;for(int j = 1; j <= N; j++) if(lx[i] + ly[j] == w[i][j] && !T[j]){T[j] = 1;if(!fa[j] || match(fa[j])){fa[j] = i;return 1;}}return 0;
}void update(){int a = INF;for(int i = 1; i <= N; i++) if(S[i])for(int j = 1; j <= N; j++) if(!T[j])a = min(a, lx[i] + ly[j] - w[i][j]);for(int i = 1; i <= N; i++){if(S[i]) lx[i] -= a;if(T[i]) ly[i] += a;}
}void KM(){for(int i = 1; i <= N; i++){fa[i] = lx[i] = ly[i] = 0;for(int j = 1; j <= N; j++) lx[i] = max(lx[i], w[i][j]);}for(int i = 1; i <= N; i++)while(1){for(int j = 1; j <= N; j++) S[j] = T[j] = 0;if(match(i)) break;else update();}
}void read(){for(int i = 1; i <= N; i++)for(int j = 1; j <= N; j++) scanf("%d", &w[i][j]);
}void solve(){for(int i = 1; i < N; i++) printf("%d ", lx[i]); printf("%d\n", lx[N]);for(int i = 1; i < N; i++) printf("%d ", ly[i]); printf("%d\n", ly[N]);int sum = 0;for(int i = 1; i <= N; i++) sum += lx[i] + ly[i];printf("%d\n", sum);
}int main()
{while(scanf("%d", &N) == 1){read();KM();solve();}return 0;
}