uva 707 - Robbery（记忆化搜索）

题目连接：uva 707 - Robbery

dp[x][y][t]表示在t时刻x，y时有没有可能存在盗贼，整个时刻只有一个点可以时就可以确定，没有点可能时就是逃脱。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>

using namespace std;
typedef pair<int,int> pii;
const int maxn = 105;
const int dir[5][2] = {{0, 1}, {0, -1}, {0, 0}, {1, 0}, {-1, 0}};

int N, M, T, dp[maxn][maxn][maxn];

void init () {
int n, t, L, X, R, Y;
memset(dp, -1, sizeof(dp));
scanf("%d", &n);
while (n--) {
scanf("%d%d%d%d%d", &t, &L, &X, &R, &Y);
for (int i = L; i <= R; i++)
for (int j = X; j <= Y; j++)
dp[i][j][t] = 0;
}
}

int dfs(int x, int y, int t) {
int& ans = dp[x][y][t];
if (ans != -1) return ans;
if (T == t) return ans = 1;
ans = 0;
for (int i = 0; i < 5; i++) {
int p = x + dir[i][0];
int q = y + dir[i][1];
if (p <= 0 || p > N || q <= 0 || q > M) continue;
if (dfs(p, q, t + 1))
ans = 1;
}
return ans;
}

vector<pii> ans[maxn];
int solve () {
for (int t = 1; t <= T; t++) {
ans[t].clear();
for (int i = 1; i <= N; i++)
for (int j = 1; j <= M; j++)
if (dp[i][j][t] == 1)
ans[t].push_back(make_pair(i, j));
if (ans[t].size() == 0) return 0;
}
bool flag = true;
for (int i = 1; i <= T; i++) {
if (ans[i].size() == 1) {
printf("Time step %d: The robber has been at %d,%d.\n", i, ans[i][0].first, ans[i][0].second);
flag = false;
}
}

return flag ? 1 : 2;
}

int main () {
int cas = 1;
while (scanf("%d%d%d", &N, &M, &T) == 3 && N + M + T) {
init ();
for (int i = 1; i <= N; i++)
for (int j = 1; j <= M; j++)
dfs(i, j, 1);
printf("Robbery #%d:\n", cas++);
int n = solve();
if (n == 0) printf("The robber has escaped.\n");
else if (n == 1) printf("Nothing known.\n");
printf("\n");
}
return 0;
}