解题思路：
二分图带权匹配，卡费用流，用KM算法。
#include <cstring>#include <algorithm>#include <cstdio>#include <iostream>using namespace std;/* KM算法 * 复杂度O(nx*nx*ny) * 求最大权匹配 * 若求最小权匹配，可将权值取相反数，结果取相反数 * 点的编号从0开始 */const int MAXN = 310;const int INF = 0x3f3f3f3f;int nx,ny;//两边的点数int g[MAXN][MAXN];//二分图描述int linker[MAXN],lx[MAXN],ly[MAXN];//y中各点匹配状态，，x,y中的点标号int slack[MAXN];bool visx[MAXN],visy[MAXN];bool DFS(int x){visx[x] = true;for(int y = 0; y < ny; y++){if(visy[y])continue;int tmp = lx[x] + ly[y] – g[x][y];if(tmp == 0){visy[y] = true;if(linker[y] == -1 || DFS(linker[y])){linker[y] = x;return true;}}else if(slack[y] > tmp)slack[y] = tmp;}return false;}int KM(){memset(linker,-1,sizeof(linker));memset(ly,0,sizeof(ly));for(int i = 0;i < nx;i++){lx[i] = -INF;for(int j = 0;j < ny;j++)if(g[i][j] > lx[i])lx[i] = g[i][j];}for(int x = 0;x < nx;x++){for(int i = 0;i < ny;i++)slack[i] = INF;while(true){memset(visx,false,sizeof(visx));memset(visy,false,sizeof(visy));if(DFS(x))break;int d = INF;for(int i = 0;i < ny;i++)if(!visy[i] && d > slack[i])d = slack[i];for(int i = 0;i < nx;i++)if(visx[i])lx[i] -= d;for(int i = 0;i < ny;i++){if(visy[i])ly[i] += d;else slack[i] -= d;}}}int res = 0;for(int i = 0;i < ny;i++)if(linker[i] != -1)res += g[linker[i]][i];return res;}int main(){int n;while(scanf("%d",&n) == 1){for(int i = 0;i < n;i++)for(int j = 0;j < n;j++)scanf("%d",&g[i][j]);nx = ny = n;printf("%d\n",KM());}return 0;}
泪，一种痛苦的雨滴，不知从什么时候开始已在我的世界下个不停。
