HDU-1853 Cyclic Tour

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1853

题目大意：

给你n个城市和m条路，现在汤姆想要旅游所有的城市，而且每个城市只能经过一次，当然，旅游路上有一点的花费，现在问汤姆怎么走才能使总花费最小。

解题思路：

二分图最优匹配的最小权问题。KM取负数实现。。经典思路。

代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<climits>using namespace std;#define N 505#define MAXN 1<<28int map[N][N];int lx[N], ly[N];int slack[N];int match[N];bool visitx[N], visity[N];int n;bool Hungary(int u){visitx[u] = true;for(int i = 1; i <= n; ++i){if(visity[i])continue;else{if(lx[u] + ly[i] == map[u][i]){visity[i] = true;if(match[i] == -1 || Hungary(match[i])){match[i] = u;return true;}}elseslack[i] = min(slack[i], lx[u] + ly[i] - map[u][i]);}}return false;}void KM_perfect_match(){int temp;for(int i = 1; i <= n; ++i)lx[i] = -MAXN;memset(ly, 0, sizeof(ly));for(int i = 1; i <= n; ++i)for(int j = 1; j <= n; ++j)lx[i] = max(lx[i], map[i][j]);for(int i = 1; i <= n; ++i){for(int j = 1; j <= n; ++j)slack[j] = MAXN;while(1){memset(visitx, false, sizeof(visitx));memset(visity, false, sizeof(visity));if(Hungary(i))break;else{temp = MAXN;for(int j = 1; j <= n; ++j)if(!visity[j])temp = min(temp, slack[j]);for(int j = 1; j <= n; ++j){if(visitx[j])lx[j] -= temp;if(visity[j])ly[j] += temp;elseslack[j] -= temp;}}}}}int main(){int m;int a, b, cost;int ans;bool flag;while(scanf("%d%d", &n, &m) != EOF){ans = 0;flag = true;memset(match, -1, sizeof(match));for(int i = 1; i <= n; ++i)for(int j = 1; j <= n; ++j)map[i][j] = -MAXN;for(int i = 1; i <= m; ++i){scanf("%d%d%d", &a, &b, &cost); //防止有重边。取较小值（负数实现）if(-cost > map[a][b])map[a][b] = -cost;}KM_perfect_match();for(int i = 1; i <= n; ++i) //是否有完美匹配{if(match[i] == -1 || map[ match[i] ][i] == -MAXN){flag = false;break;}ans += map[match[i]][i];}if(flag)printf("%d\n", -ans);elseprintf("-1\n");}return 0;}