hdu2389 Rain on your Parade--HK算法 & 最大匹配数

来源：互联网发布：台湾知见旅行社编辑：程序博客网时间：2024/06/04 19:51

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=2389

题意：pNum个人，uNum把伞，给定这些人和伞的二维坐标，和每个人的行进速度。问在t时间内，最多有多少人能拿到伞？

建二分图，左边是人，与右边是伞，如果这个人到伞的时间不大于t就连线，最后求最大匹配就行，不过这题用匈牙利算法会超时，所以用HK算法。

#define _CRT_SECURE_NO_DEPRECATE #include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<queue>#include<vector>#include<string>#define INF 99999999;using namespace std;struct People{double x;double y;double t;};struct Umbrella{double x;double y;};int T;int time;//约束时间int pNum;//人的数目int uNum;//伞的数目People people[3005];Umbrella umbrella[3005];int dis;int cx[3005];int cy[3005];int dx[3005];int dy[3005];bool g[3005][3005];bool vis[3005];double distance(People p, Umbrella u){return sqrt((p.x - u.x)*(p.x - u.x) + (p.y - u.y)*(p.y - u.y));}bool searchPath(){dis = INF;queue<int> Q;memset(dx, -1, sizeof(dx));memset(dy, -1, sizeof(dy));for (int i = 0; i < pNum; i++){if (cx[i] == -1){Q.push(i);dx[i] = 0;}}while (!Q.empty()){int u = Q.front();Q.pop();if (dx[u] > dis)break;for (int v = 0; v < uNum; v++){if (g[u][v] && dy[v] == -1){dy[v] = dx[u] + 1;if (cy[v] == -1)dis = dy[v];else{dx[cy[v]] = dy[v] + 1;Q.push(cy[v]);}}}}return dis != INF;}int dfs(int u){for (int v = 0; v < uNum; v++){if (g[u][v] && dy[v] == dx[u] + 1 && !vis[v]){vis[v] = 1;if (cy[v] != -1 && dy[v] == dis)continue;if (cy[v] == -1 || dfs(cy[v])){cx[u] = v;cy[v] = u;return 1;}}}return 0;}int HK(){int ans = 0;memset(cx, -1, sizeof(cx));memset(cy, -1, sizeof(cy));while (searchPath()){memset(vis, 0, sizeof(vis));for (int u = 0; u < pNum; u++){if (cx[u] == -1)ans += dfs(u);}}return ans;}int main(){scanf("%d", &T);int cas = 1;while (T--){scanf("%d%d", &time, &pNum);for (int i = 0; i < pNum; i++)scanf("%lf%lf%lf", &people[i].x, &people[i].y, &people[i].t);scanf("%d", &uNum);for (int i = 0; i < uNum; i++)scanf("%lf%lf", &umbrella[i].x, &umbrella[i].y);memset(g, 0, sizeof(g));for (int i = 0; i < pNum; i++)for (int j = 0; j < uNum; j++)if (distance(people[i], umbrella[j]) / people[i].t <= time)g[i][j] = 1;printf("Scenario #%d:\n%d\n\n", cas++, HK());}return 0;}

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