POJ 3592 Instantaneous Transference

题目链接

题意：一个图。能往右和下走，然后有*能够传送到一个位置。'#'不能走。走过一个点能够获得该点上面的数字值，问最大能获得多少

思路：因为有环先强连通缩点。然后问题转化为dag，直接dp就可以

代码：

#include <cstdio>

#include <cstring>

#include <vector>

#include <algorithm>

#include <stack>

using namespace std;

const int N = 1605;

const int d[2][2] = {0, 1, 1, 0};

int t, n, m, val[N];

char str[45][45];

vector<int> g[N], scc[N];

stack<int> S;

int pre[N], dfn[N], dfs_clock, sccno[N], sccn, scc_val[N];

void dfs_scc(int u) {

	pre[u] = dfn[u] = ++dfs_clock;

	S.push(u);

	for (int i = 0; i < g[u].size(); i++) {

		int v = g[u][i];

		if (!pre[v]) {

			dfs_scc(v);

			dfn[u] = min(dfn[u], dfn[v]);

		} else if (!sccno[v]) dfn[u] = min(dfn[u], pre[v]);

	}

	if (dfn[u] == pre[u]) {

		sccn++;

		int sum = 0;

		while (1) {

			int x = S.top(); S.pop();

			sum += val[x];

			sccno[x] = sccn;

			if (u == x) break;

		}

		scc_val[sccn] = sum;

	}

}

void find_scc() {

	dfs_clock = sccn = 0;

	memset(pre, 0, sizeof(pre));

	memset(sccno, 0, sizeof(sccno));

	for (int i = 0; i < n * m; i++)

		if (!pre[i]) dfs_scc(i);

}

int dp[N];

int dfs(int u) {

	if (dp[u] != -1) return dp[u];

	dp[u] = 0;

	for (int i = 0; i < scc[u].size(); i++) {

		int v = scc[u][i];

		dp[u] = max(dp[u], dfs(v));

	}

	dp[u] += scc_val[u];

	return dp[u];

}

int main() {

	scanf("%d", &t);

	while (t--) {

		scanf("%d%d", &n, &m);

		for (int i = 0; i < n * m; i++) g[i].clear();

		for (int i = 0; i < n; i++)

			scanf("%s", str[i]);

		memset(val, 0, sizeof(val));

		int a, b;

		for (int i = 0; i < n; i++) {

			for (int j = 0; j < m; j++) {

				if (str[i][j] == '#') continue;

				if (str[i][j] >= '0' && str[i][j] <= '9') val[i * m + j] = str[i][j] - '0';

				if (str[i][j] == '*') {

					scanf("%d%d", &a, &b);

					g[i * m + j].push_back(a * m + b);

				}

				for (int k = 0; k < 2; k++) {

					int x = i + d[k][0];

					int y = j + d[k][1];

					if (x < 0 || x >= n || y < 0 || y >= m || str[x][y] == '#') continue;

					g[i * m + j].push_back(x * m + y);

				}

			}

		}

		find_scc();

		for (int i = 1; i <= sccn; i++) scc[i].clear();

		for (int u = 0; u < n * m; u++) {

			for (int j = 0; j < g[u].size(); j++) {

				int v = g[u][j];

				if (sccno[u] == sccno[v]) continue;

				scc[sccno[u]].push_back(sccno[v]);

			}

		}

		memset(dp, -1, sizeof(dp));

		printf("%d\n", dfs(sccno[0]));

	}

	return 0;

}