贪心+网络流。对于每个结点,构建入点和出点。
对于每一个lizard>0,构建边s->in position of lizard, 容量为1.
对于pillar>0, 构建边in position of pillar -> out position of pillar, 容量为number of pillar.
若沿四个方向移动距离d可以超过边界,则构建边out position of pillar -> t, 容量为INF;
否则, 对于曼哈顿距离l(l>0 and l<=d)的点p构建边out position of pillar -> in position p, 容量为INF。(此时,不用考虑p是否含有pillar,无所谓)
Dinic可解网络流。一定要注意输出was/were,单复数和0等。
1 /* 2732 */ 2 #include <iostream> 3 #include <string> 4 #include <map> 5 #include <queue> 6 #include <set> 7 #include <stack> 8 #include <vector> 9 #include <deque> 10 #include <algorithm> 11 #include <cstdio> 12 #include <cmath> 13 #include <ctime> 14 #include <cstring> 15 #include <climits> 16 #include <cctype> 17 #include <cassert> 18 #include <functional> 19 #include <iterator> 20 #include <iomanip> 21 using namespace std; 22 //#pragma comment(linker,"/STACK:102400000,1024000") 23 24 #define sti set<int> 25 #define stpii set<pair<int, int> > 26 #define mpii map<int,int> 27 #define vi vector<int> 28 #define pii pair<int,int> 29 #define vpii vector<pair<int,int> > 30 #define rep(i, a, n) for (int i=a;i<n;++i) 31 #define per(i, a, n) for (int i=n-1;i>=a;--i) 32 #define clr clear 33 #define pb push_back 34 #define mp make_pair 35 #define fir first 36 #define sec second 37 #define all(x) (x).begin(),(x).end() 38 #define SZ(x) ((int)(x).size()) 39 #define lson l, mid, rt<<1 40 #define rson mid+1, r, rt<<1|1 41 42 const int INF = 0x1f1f1f1f; 43 const int maxn = 25; 44 const int maxv = maxn*maxn*2; 45 const int maxe = maxv*maxn*8; 46 int V[maxe], F[maxe], nxt[maxe]; 47 int head[maxv], dis[maxv]; 48 char Ps[maxn][maxn], Ls[maxn][maxn]; 49 int dir[4][2] = { 50 -1, 0, 1, 0, 0, -1, 0, 1 51 }; 52 int n, m; 53 54 void addEdge(int u, int v, int c) { 55 V[m] = v; 56 F[m] = c; 57 nxt[m] = head[u]; 58 head[u] = m++; 59 60 V[m] = u; 61 F[m] = 0; 62 nxt[m] = head[v]; 63 head[v] = m++; 64 } 65 66 bool bfs(int s, int t) { 67 queue<int> Q; 68 int u, v, k; 69 70 memset(dis, 0, sizeof(dis)); 71 Q.push(s); 72 dis[s] = 1; 73 74 while (!Q.empty()) { 75 u = Q.front(); 76 Q.pop(); 77 for (k=head[u]; k!=-1; k=nxt[k]) { 78 v = V[k]; 79 if (!dis[v] && F[k]) { 80 dis[v] = dis[u] + 1; 81 Q.push(v); 82 } 83 } 84 } 85 86 return dis[t]==0; 87 } 88 89 int dfs(int u, int t, int val) { 90 if (u==t || val==0) 91 return val; 92 93 int ret = 0; 94 int tmp, v, k; 95 96 for (k=head[u]; k!=-1; k=nxt[k]) { 97 v = V[k]; 98 if (dis[v]==dis[u]+1 && F[k] && (tmp=dfs(v, t, min(val, F[k])))>0) { 99 F[k] -= tmp; 100 F[k^1] += tmp; 101 ret += tmp; 102 val -= tmp; 103 if (val == 0) 104 break; 105 } 106 } 107 108 return ret; 109 } 110 111 int Dinic(int s, int t) { 112 int ret = 0, tmp; 113 114 while (1) { 115 if (bfs(s, t)) 116 break; 117 118 tmp = dfs(s, t, INF); 119 ret += tmp; 120 } 121 122 return ret; 123 } 124 125 int main() { 126 ios::sync_with_stdio(false); 127 #ifndef ONLINE_JUDGE 128 freopen("data.in", "r", stdin); 129 freopen("data.out", "w", stdout); 130 #endif 131 132 int case_n; 133 int s, t, n_, m_, d, tmp; 134 int x, y; 135 int k, id, id_; 136 int ans, tot; 137 bool flag; 138 139 scanf("%d", &case_n); 140 rep(tt, 1, case_n+1) { 141 scanf("%d %d", &n_, &d); 142 143 rep(i, 1, n_+1) 144 scanf("%s", Ps[i]+1); 145 rep(i, 1, n_+1) 146 scanf("%s", Ls[i]+1); 147 148 m_ = strlen(Ps[1]+1); 149 tot = 0; 150 s = m = 0; 151 k = n_ * m_; 152 t = k*2+1; 153 memset(head, -1, sizeof(head)); 154 155 rep(i, 1, n_+1) { 156 rep(j, 1, m_+1) { 157 id = (i-1) * m_ + j; 158 if (Ls[i][j] == ‘L‘) { 159 addEdge(s, id, 1); 160 ++tot; 161 } 162 if (Ps[i][j] > ‘0‘) { 163 addEdge(id, id+k, Ps[i][j]-‘0‘); 164 flag = true; 165 id += k; 166 rep(kk, 0, 4) { 167 x = i + dir[kk][0] * d; 168 y = j + dir[kk][1] * d; 169 if (x<=0 || x>n_ || y<=0 || y>m_) { 170 addEdge(id, t, INF); 171 flag = false; 172 break; 173 } 174 } 175 if (flag) { 176 id_ = 0; 177 rep(ii, 1, n_+1) { 178 rep(jj, 1, m_+1) { 179 ++id_; 180 tmp = abs(ii-i) + abs(jj-j); 181 if (tmp>0 && tmp<=d) 182 addEdge(id, id_, Ps[i][j]-‘0‘); 183 } 184 } 185 } 186 } 187 } 188 } 189 190 ans = tot - Dinic(s, t); 191 printf("Case #%d: ", tt); 192 if (ans == 0) { 193 printf("no lizard was left behind.\n"); 194 } else if (ans == 1) { 195 printf("1 lizard was left behind.\n"); 196 } else { 197 printf("%d lizards were left behind.\n", ans); 198 } 199 } 200 201 #ifndef ONLINE_JUDGE 202 printf("time = %d.\n", (int)clock()); 203 #endif 204 205 return 0; 206 }
【HDOJ】2732 Leapin' Lizards
时间: 2024-12-26 09:29:16