【圖論02】二分圖 1002 奔小康賺大錢

演算法思路：網路流， 模板題。

#include<iostream>#include<cstdio>#include<cstring>#include<climits>#include<algorithm>using namespace std;#define N 310int map[N][N];bool visitx[N], visity[N];int lx[N], ly[N];int match[N];int n;bool Hungary(int u) //匈牙利演算法{visitx[u] = true;for(int i = 0; i < n; ++i){if(!visity[i] && lx[u] + ly[i] == map[u][i]){visity[i] = true;if(match[i] == -1 || Hungary(match[i])){match[i] = u;return true;}}}return false;}void KM_perfect_match(){int temp;memset(lx, 0, sizeof(lx)); //初始化頂標memset(ly, 0, sizeof(ly)); //ly[i]為0for(int i = 0; i < n; ++i) //lx[i]為權值最大的邊for(int j = 0; j < n; ++j)lx[i] = max(lx[i], map[i][j]);for(int i = 0; i < n; ++i) //對n個點匹配{while(1){memset(visitx, false, sizeof(visitx));memset(visity, false, sizeof(visity));if(Hungary(i)) //匹配成功break;else //匹配失敗，找最小值{temp = INT_MAX;for(int j = 0; j < n; ++j) //x在交錯樹中if(visitx[j])for(int k = 0; k < n; ++k) //y在交錯樹外if(!visity[k] && temp > lx[j] + ly[k] - map[j][k])temp = lx[j] + ly[k] - map[j][k];for(int j = 0; j < n; ++j) //更新頂標{if(visitx[j])lx[j] -= temp;if(visity[j])ly[j] += temp;}}}}}int main(){int ans;while(scanf("%d", &n) != EOF){ans = 0;memset(match, -1, sizeof(match));for(int i = 0; i < n; ++i)for(int j = 0; j < n; ++j)scanf("%d", &map[i][j]);KM_perfect_match();for(int i = 0; i < n; ++i) //權值相加ans += map[match[i]][i];printf("%d\n", ans);}return 0;}

轉載自：http://blog.csdn.net/niushuai666/article/details/7171525