BZOJ 2464 中山市选 2009 小明的游戏 最短路

题目大意：给出一个地图，如果经过两个不同的区块，需要花费1，否则不需要花费。问从st到ed最小需要花费多少。

思路：签到题。

CODE：

#include <queue>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 510#define MAXP 250010#define MAXE 2000010using namespace std;const int dx[] = {0,1,-1,0,0};const int dy[] = {0,0,0,1,-1};struct Status{int pos,len;Status(int _,int __):pos(_),len(__) {}bool operator <(const Status &a)const {return len > a.len;}};char src[MAX][MAX];int num[MAX][MAX];int head[MAXP],total;int next[MAXE],aim[MAXE],length[MAXE];int st,ed,m,n;int f[MAXP];inline void Initialize(){total = 0;memset(head,0,sizeof(head));}inline void Add(int x,int y,int len){next[++total] = head[x];aim[total] = y;length[total] = len;head[x] = total;}inline int SPFA(){static priority_queue<Status> q;while(!q.empty())q.pop();memset(f,0x3f,sizeof(f));q.push(Status(st,0));f[st] = 0;while(!q.empty()) {Status now = q.top(); q.pop();if(f[now.pos] < now.len)continue;int x = now.pos;for(int i = head[x]; i; i = next[i])if(f[aim[i]] > f[x] + length[i]) {f[aim[i]] = f[x] + length[i];q.push(Status(aim[i],f[aim[i]]));}}return f[ed];}int main(){while(scanf("%d%d",&m,&n),m + n) {Initialize();for(int i = 1; i <= m; ++i)scanf("%s",src[i] + 1);int cnt = 0;for(int i = 1; i <= m; ++i)for(int j = 1; j <= n; ++j)num[i][j] = ++cnt;int x,y;scanf("%d%d",&x,&y);x++,y++,st = num[x][y];scanf("%d%d",&x,&y);x++,y++,ed = num[x][y];for(int i = 1; i <= m; ++i)for(int j = 1; j <= n; ++j)for(int k = 1; k <= 4; ++k) {int fx = i + dx[k];int fy = j + dy[k];if(!fx * fy || fx > m || fy > n)continue;Add(num[i][j],num[fx][fy],src[fx][fy] != src[i][j]);}printf("%d\n",SPFA());}return 0;}

CODE：