[BZOJ1001][BeiJing2006]狼抓兔子
试题描述
输入
输出
输出一个整数,表示参与伏击的狼的最小数量.
输入示例
3 4 5 6 4 4 3 1 7 5 3 5 6 7 8 8 7 6 5 5 5 5 6 6 6
输出示例
14
数据规模及约定
见“输入”
题解
转化成对偶图跑最短路特别快。经典问题了。
这题强行出 n = 1 或 m = 1 的数据,真是 ****
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <algorithm> #include <queue> using namespace std; int read() { int x = 0, f = 1; char c = getchar(); while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); } while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); } return x * f; } #define maxn 1010 #define maxnode 2000010 #define maxm 6000010 #define oo 2147483647 int N, M, head[maxnode], nxt[maxm], to[maxm], dist[maxm]; void AddEdge(int a, int b, int c) { to[++M] = b; dist[M] = c; nxt[M] = head[a]; head[a] = M; swap(a, b); to[++M] = b; dist[M] = c; nxt[M] = head[a]; head[a] = M; return ; } int upt(int x, int y, int m) { return ((x - 1) * (m - 1) + y << 1) - 1; } int dnt(int x, int y, int m) { return upt(x, y, m) + 1; } struct Node { int u, d; Node() {} Node(int _, int __): u(_), d(__) {} bool operator < (const Node& t) const { return d > t.d; } }; priority_queue <Node> Q; int d[maxnode]; bool vis[maxnode]; void solve(int s, int t) { memset(vis, 0, sizeof(vis)); for(int i = 1; i <= N; i++) d[i] = oo; while(!Q.empty()) Q.pop(); d[s] = 0; Q.push(Node(s, 0)); while(!Q.empty()) { int u = Q.top().u; Q.pop(); if(vis[u]) continue; vis[u] = 1; if(u == t){ printf("%d\n", d[u]); break; } for(int e = head[u]; e; e = nxt[e]) if(d[to[e]] > d[u] + dist[e]) { d[to[e]] = d[u] + dist[e]; if(!vis[to[e]]) Q.push(Node(to[e], d[to[e]])); } } return ; } int main() { int n, m; scanf("%d%d", &n, &m); if(m == 1) { int mn = oo; for(int i = 1; i < n; i++) mn = min(mn, read()); return printf("%d\n", mn), 0; } if(n == 1) { int mn = oo; for(int i = 1; i < m; i++) mn = min(mn, read()); return printf("%d\n", mn), 0; } M = 0; memset(head, 0, sizeof(head)); N = ((n - 1) * (m - 1) << 1) + 2; int s = N - 1, t = N; for(int i = 1; i <= n; i++) for(int j = 1; j < m; j++) { int u, v; if(i == 1) { u = upt(i, j, m); AddEdge(s, u, read()); } if(1 < i && i < n) { u = upt(i, j, m); v = dnt(i - 1, j, m); AddEdge(u, v, read()); } if(i == n) { u = dnt(i - 1, j, m); AddEdge(u, t, read()); } } for(int i = 1; i < n; i++) for(int j = 1; j <= m; j++) { int u, v; if(j == 1) { u = dnt(i, j, m); AddEdge(u, t, read()); } if(1 < j && j < m) { u = upt(i, j - 1, m); v = dnt(i, j, m); AddEdge(u, v, read()); } if(j == m) { u = upt(i, j - 1, m); AddEdge(s, u, read()); } } for(int i = 1; i < n; i++) for(int j = 1; j < m; j++) { int u = upt(i, j, m), v = dnt(i, j, m); AddEdge(u, v, read()); } solve(s, t); return 0; }
原文:http://www.cnblogs.com/xiao-ju-ruo-xjr/p/6498024.html