題目鏈接

Problem Description

Your platoon of wandering lizards has entered a strange room in the labyrinth you are exploring. As you are looking around for hidden treasures, one of the rookies steps on an innocent-looking stone and the room’s floor suddenly disappears! Each lizard in your platoon is left standing on a fragile-looking pillar, and a fire begins to rage below… Leave no lizard behind! Get as many lizards as possible out of the room, and report the number of casualties.The pillars in the room are aligned as a grid, with each pillar one unit away from the pillars to its east, west, north and south. Pillars at the edge of the grid are one unit away from the edge of the room (safety). Not all pillars necessarily have a lizard. A lizard is able to leap onto any unoccupied pillar that is within d units of his current one. A lizard standing on a pillar within leaping distance of the edge of the room may always leap to safety… but there’s a catch: each pillar becomes weakened after each jump, and will soon collapse and no longer be usable by other lizards. Leaping onto a pillar does not cause it to weaken or collapse; only leaping off of it causes it to weaken and eventually collapse. Only one lizard may be on a pillar at any given time.

Input

The input file will begin with a line containing a single integer representing the number of test cases, which is at most 25. Each test case will begin with a line containing a single positive integer n representing the number of rows in the map, followed by a single non-negative integer d representing the maximum leaping distance for the lizards. Two maps will follow, each as a map of characters with one row per line. The first map will contain a digit (0-3) in each position representing the number of jumps the pillar in that position will sustain before collapsing (0 means there is no pillar there). The second map will follow, with an ‘L’ for every position where a lizard is on the pillar and a ‘.’ for every empty pillar. There will never be a lizard on a position where there is no pillar.Each input map is guaranteed to be a rectangle of size n x m, where 1 ≤ n ≤ 20 and 1 ≤ m ≤ 20. The leaping distance isalways 1 ≤ d ≤ 3.

Output

For each input case, print a single line containing the number of lizards that could not escape. The format should follow the samples provided below.

Sample Input

4 3 1 1111 1111 1111 LLLL LLLL LLLL 3 2 00000 01110 00000 ..... .LLL. ..... 3 1 00000 01110 00000 ..... .LLL. ..... 5 2 00000000 02000000 00321100 02000000 00000000 ........ ........ ..LLLL.. ........ ........

Sample Output

Case #1: 2 lizards were left behind. Case #2: no lizard was left behind. Case #3: 3 lizards were left behind. Case #4: 1 lizard was left behind.

AC

• 建圖

• 源點到蜥蜴所在的位置編號建邊，流量為1
• 遍歷圖上每個點，把所有可以相互跳躍的點建邊，流量為inf
• 途中的每個點拆點建邊，流量為可以跳躍的次數
• 邊緣的點到匯點建邊，流量為inf

• 注意輸出：were，was，lizard，lizards

#include <iostream> #include <stdio.h> #include <map> #include <vector> #include <queue> #include <algorithm> #include <cmath> #define N 1005 #include <cstring> #define ll long long #define P pair<int, int> #define mk make_pair using namespace std; struct ac{int v, c, pre; }edge[800001]; int head[N], dis[N], curedge[N], cnt; int inf = 0x3f3f3f3f; void addedge(int u, int v, int c) {edge[cnt].v = v;edge[cnt].c = c;edge[cnt].pre = head[u];head[u] = cnt++;swap(u, v);edge[cnt].v = v;edge[cnt].c = 0;edge[cnt].pre = head[u];head[u] = cnt++; } bool bfs(int s, int e) {queue<int> que;que.push(s);memset(dis, 0, sizeof(dis));dis[s] = 1;while (!que.empty()) {int t = que.front();que.pop();for (int i = head[t]; i != -1; i = edge[i].pre) {if (dis[edge[i].v] || edge[i].c == 0) continue;dis[edge[i].v] = dis[t] + 1;que.push(edge[i].v);} }return dis[e] != 0; } int dfs(int s, int e, int flow) {if (s == e) return flow;for (int &i = curedge[s]; i != -1; i = edge[i].pre) {if (dis[edge[i].v] == dis[s] + 1 && edge[i].c) {int d = dfs(edge[i].v, e, min(flow, edge[i].c));if (d > 0) {edge[i].c -= d;edge[i ^ 1].c += d;return d; }}}return 0; } int n, d, m; int solve(int s, int e) {int sum = 0;while (bfs(s, e)) {for (int i = 0; i <= e; ++i) {curedge[i] = head[i];}int d;while (d = dfs(s, e, inf)) {sum += d;}}return sum; } char map1[40][40], map2[40][40];bool vis_map[40][40]; bool judge_pos(int x, int y) {if (x < 0 || y < 0 || x >= n || y >= m) return false;return true; } int dist(int x1, int y1, int x2, int y2) {return (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2); } bool judge_edge(int x, int y) {if (x < d) return true;if (n - x - 1 < d) return true;if (y < d) return true;if (m - y - 1 < d) return true;return false; } int main() { #ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin); #endifint t;scanf("%d", &t);int Case = 1;while (t--) {memset(head, -1, sizeof(head));cnt = 0;// 讀入兩個地圖 scanf("%d%d", &n, &d);for (int i = 0; i < n; ++i) {scanf("%s", map1[i]);}for (int i = 0; i < n; ++i) {scanf("%s", map2[i]);}m = strlen(map1[0]);// 源點到蜥蜴建邊，流量為1 int start = 0, end = n * m * 2 + 1, sum = 0;for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {if (map2[i][j] == 'L') {sum++;addedge(start, i * m + j + 1, 1);}}}// 對 map1 拆點建圖， 權值為可以跳躍的次數 for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {int spos = i * m + (j + 1);int epos = spos + n * m;int num = map1[i][j] - '0';addedge(spos, epos, num);}}// 枚舉地圖中的位置，將所有可以相互跳躍的點建邊， 權值為inf for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {for (int k = -d; k <= d; ++k) {for (int z = -d; z <= d; ++z) {int x = i + k;int y = j + z; if (x == i && y == j) continue;// fabs(i - x) + fabs(j - y) <= d // 判斷下個位置是否能跳過去 if (judge_pos(x, y) && dist(x, y, i, j) <= d * d) {int s = i * m + j + 1;int e = x * m + y + 1;addedge(s + n * m, e, inf);}}}}}// 邊緣的點到匯點建邊 for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {if (judge_edge(i, j)) {addedge(i * m + j + 1 + n * m, end, inf);}}}int ans = solve(start, end);// 注意輸出，were，was，lizard，lizards if (sum - ans == 0) printf("Case #%d: no lizard was left behind.\n", Case++);else if (sum - ans == 1) printf("Case #%d: %d lizard was left behind.\n", Case++, sum - ans);else printf("Case #%d: %d lizards were left behind.\n", Case++, sum - ans);} return 0; }

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