【BZOJ2464】【中山市选2009】小明的游戏 最短路水过

题解：最短路pqspfa200ms，一眼题，

另一种想出来没写的做法：二分答案，上界n+m

时间复杂度O(n*m*log(n+m))，二分+深搜看能不能找到t

最短路代码：

#include <queue>#include <cstdio>#include <cstring>#include <algorithm>#define N 505#define NN 251000#define inf 0x3f3f3f3fusing namespace std;const int dx[]={0,0,1,-1};const int dy[]={1,-1,0,0};struct KSD{int v,len,next;}e[NN<<2];int head[NN],cnt;void add(int u,int v,int len){++cnt;e[cnt].v=v;e[cnt].len=len;e[cnt].next=head[u];head[u]=cnt;}int dist[NN],in[NN];struct Vayne{int f,v;bool operator < (const Vayne &a)const{return f>a.f;}Vayne(){}Vayne(int _f,int _v):f(_f),v(_v){}};int s,t,n,m,id[N][N],num;char map[N][N];priority_queue<Vayne>q;void init(){num=cnt=0;memset(head,0,sizeof(head));}int spfa(){while(!q.empty())q.pop();memset(dist,0x3f,sizeof(dist));memset(in,0,sizeof(in));int i,u,v,temp;dist[s]=0;in[s]=1;q.push(Vayne(0,s));while(!q.empty()){Vayne U=q.top();q.pop();u=U.v;in[u]=0;for(i=head[u];i;i=e[i].next){v=e[i].v;if(dist[v]>dist[u]+e[i].len){dist[v]=dist[u]+e[i].len;if(!in[v]){in[v]=1;q.push(Vayne(dist[v],v));}}}}return dist[t];}bool build(){int i,j,k;int a,b,c;int x,y;init();scanf("%d%d",&n,&m);if(n==0&&m==0)return 0;for(i=1;i<=n;i++)scanf("%s",map[i]+1);for(i=1;i<=n;i++)for(j=1;j<=m;j++)id[i][j]=++num;for(i=1;i<=n;i++)for(j=1;j<=m;j++){int iid=id[i][j];for(k=0;k<4;k++){x=i+dx[k];y=j+dy[k];if(id[x][y]){int len=(map[i][j]==map[x][y]);add(iid,id[x][y],!len);add(id[x][y],iid,!len);}}}scanf("%d%d%d%d",&a,&b,&x,&y);s=id[a+1][b+1];t=id[x+1][y+1];return 1;}int main(){//freopen("test.in","r",stdin);while(build())printf("%d\n",spfa());return 0;}