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读题显然是二分图匹配,看成guest与umbrella的匹配。匈牙利果断TLE了,其实时间卡的相当紧。HK过的,750ms。
1 /* 2389 */ 2 #include <iostream> 3 #include <string> 4 #include <map> 5 #include <queue> 6 #include <set> 7 #include <stack> 8 #include <vector> 9 #include <algorithm> 10 #include <cstdio> 11 #include <cmath> 12 #include <ctime> 13 #include <cstring> 14 #include <climits> 15 #include <cctype> 16 using namespace std; 17 18 typedef struct { 19 int v, next; 20 } Edge_t; 21 22 const int maxn = 3005; 23 24 Edge_t E[maxn*maxn]; 25 int head[maxn], L; 26 int dx[maxn], dy[maxn]; 27 int xpre[maxn], ypre[maxn]; 28 int gx[maxn], gy[maxn], gs[maxn]; 29 bool visit[maxn]; 30 int rt, n, m; 31 int ans, dis; 32 33 void init() { 34 L = 0; 35 memset(head, -1, sizeof(head)); 36 memset(xpre, -1, sizeof(xpre)); 37 memset(ypre, -1, sizeof(ypre)); 38 } 39 40 void addEdge(int u, int v) { 41 E[L].v = v; 42 E[L].next = head[u]; 43 head[u] = L++; 44 } 45 46 bool bfs() { 47 int i, j, k; 48 int u, v; 49 queue<int> Q; 50 51 memset(dx, -1, sizeof(dx)); 52 memset(dy, -1, sizeof(dy)); 53 dis = INT_MAX; 54 55 for (i=0; i<m; ++i) { 56 if (xpre[i] == -1) { 57 Q.push(i); 58 dx[i] = 0; 59 } 60 } 61 62 while (!Q.empty()) { 63 u = Q.front(); 64 Q.pop(); 65 if (dx[u] > dis) 66 break; 67 for (i=head[u]; i!=-1; i=E[i].next) { 68 v = E[i].v; 69 if (dy[v] == -1) { 70 dy[v] = dx[u] + 1; 71 if (ypre[v] == -1) { 72 dis = dy[v]; 73 } else { 74 dx[ypre[v]] = dy[v] + 1; 75 Q.push(ypre[v]); 76 } 77 } 78 } 79 } 80 81 return dis!=INT_MAX; 82 } 83 84 int dfs(int u) { 85 int i, v; 86 // Edge_t e; 87 88 for (i=head[u]; i!=-1; i=E[i].next) { 89 v = E[i].v; 90 if (!visit[v] && dy[v]==dx[u]+1) { 91 visit[v] = true; 92 if (ypre[v]!=-1 && dy[v]==dis) 93 continue; 94 if (ypre[v]==-1 || dfs(ypre[v])) { 95 xpre[u] = v; 96 ypre[v] = u; 97 return 1; 98 } 99 } 100 } 101 102 return 0; 103 } 104 105 int HK() { 106 int ret = 0; 107 108 while (bfs()) { 109 memset(visit, false, sizeof(visit)); 110 for (int i=0; i<m; ++i) 111 if (xpre[i] == -1) 112 ret += dfs(i); 113 } 114 return ret; 115 } 116 117 int main() { 118 int i, j, k; 119 int t, tt; 120 int x, y; 121 122 #ifndef ONLINE_JUDGE 123 freopen("data.in", "r", stdin); 124 freopen("data.out", "w", stdout); 125 #endif 126 127 scanf("%d", &tt); 128 for (t=1; t<=tt; ++t) { 129 scanf("%d", &rt); 130 scanf("%d", &m); 131 init(); 132 for (i=0; i<m; ++i) 133 scanf("%d %d %d", &gx[i], &gy[i], &gs[i]); 134 scanf("%d", &n); 135 for (i=0; i<n; ++i) { 136 scanf("%d %d", &x, &y); 137 for (j=0; j<m; ++j) { 138 dis = gs[i] * rt; 139 if ((x-gx[j])*(x-gx[j])+(y-gy[j])*(y-gy[j]) <= dis*dis) 140 addEdge(j, i); 141 } 142 } 143 ans = HK(); 144 printf("Scenario #%d:\n%d\n\n", t, ans); 145 } 146 147 #ifndef ONLINE_JUDGE 148 printf("%d\n", (int)clock()); 149 #endif 150 151 return 0; 152 }
【HDOJ】2389 Rain on your Parade
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原文地址:http://www.cnblogs.com/bombe1013/p/4432367.html