标签：style 表示 stdout logs scan usaco can pen return
题意
有一幅n*n的方格图，n <= 100，每个点上有一个值。从（1,1）出发，走到（n,n），只能走四联通。每走一步花费t，每走三步需要花费走完三步后到达格子的值。求最小花费的值。
拆点，dis[i][j]表示到达第i个点时走的总步数模3等于j时的最小花费值。
#include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <algorithm> #include <iostream> #include <queue> using namespace std; const int maxn = 105; const int INF = 0x3fffffff; int n, t, w[maxn][maxn]; struct Node { int x, y, z; Node (int x = 0, int y = 0, int z = 0): x(x), y(y), z(z) {} }; queue <Node> q; int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0}; int dis[maxn][maxn][3]; bool vis[maxn][maxn][3]; int main() { freopen("visitfj.in", "r", stdin); freopen("visitfj.out", "w", stdout); scanf("%d %d", &n, &t); for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) scanf("%d", &w[i][j]); for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) for (int k = 0; k < 3; ++k) dis[i][j][k] = INF, vis[i][j][k] = false; dis[1][1][0] = 0; vis[1][1][0] = true; q.push(Node(1, 1, 0)); while (!q.empty()) { Node now = q.front(); vis[now.x][now.y][now.z] = false; q.pop(); for (int i = 0; i < 4; ++i) { int x = now.x+dx[i], y = now.y+dy[i], z = (now.z+1)%3; if (x < 1 || y < 1 || x > n || y > n) continue ; if (dis[x][y][z] > dis[now.x][now.y][now.z]+t+((z == 0) ? w[x][y] : 0)) { dis[x][y][z] = dis[now.x][now.y][now.z]+t+((z == 0) ? w[x][y] : 0); if (!vis[x][y][z]) vis[x][y][z] = true, q.push(Node(x, y, z)); } } } printf("%d\n", min(dis[n][n][0], min(dis[n][n][1], dis[n][n][2]))); return 0; }
USACO 2017 FEB Gold visitfj 最短路
标签：style 表示 stdout logs scan usaco can pen return
原文地址：http://www.cnblogs.com/-ZZB-/p/6403392.html
