ZOJ3781 - Paint the Grid Reloaded(最短路)

题目链接 ZOJ3781

【题意】给出n*m矩阵，每一格有黑白两种颜色，相邻的同一种颜色的联通快可以一起翻转成另外一种颜色，求所有块变成同一种颜色最少需要翻转几次。

【分析】这题是4月份省赛上留下来的，一开始用了最无脑的bfs，把每一个矩阵保存成状态压入队列，结果超内存。然后后来看了别人题解才知道可以看成最短路问题。先用dfs给所有连同块标号，顺便把相邻的不同颜色的连通块连上边，然后怎样看出来是最短路呢，我想了很久。 其实可以看成枚举从任意一个连通块出发，能走到的最远的连通块就是本次的答案，然后取所有最小值即可，可以把这看成从一块连通块出发，向水的扩散一样，并且向四方扩散速度相同，这样找到最远到达的那一块就是把所有快变成同一颜色的步数。

至于求最短路，我写了两种，优化过的dijkstra和bfs

【AC CODE(dijkstra)】430ms

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;#define MAXN 45#define INF 0x3f3f3f3f#define hash(a,b) (a*w+b)#define max(a,b) (a>b?a:b)struct NODE{int dis, num;NODE(int a = 0, int b = 0){dis = a, num = b;}bool operator <(const NODE& a) const {return dis > a.dis;}};const int d[4][2]={{-1,0},{1,0},{0,-1},{0,1}};char map[MAXN][MAXN];int h,w, f[MAXN*MAXN], cnt, dp[MAXN*MAXN];vector<int> g[MAXN*MAXN];bool vis[MAXN*MAXN];priority_queue<NODE> q;void dfs(int si, int sj){f[hash(si,sj)] = cnt;for(int i = 0; i < 4; i++){int ni = d[i][0]+si, nj = d[i][1]+sj;if(0 <= ni && h > ni && 0 <= nj && w > nj){if(map[ni][nj] == map[si][sj]){if(-1 == f[hash(ni,nj)])dfs(ni,nj);}else if(~f[hash(ni,nj)]){int num = f[hash(ni,nj)];g[cnt].push_back(num);g[num].push_back(cnt);}}}}int dijkstra(int st){memset(dp,0x3f,sizeof(dp));memset(vis,0,sizeof(vis));dp[st] = 0;q.push(NODE(0,st));while(!q.empty()){NODE now = q.top();q.pop();int u = now.num;if(vis[u]) continue;vis[u] = true;int len = g[u].size();for(int i = 0; i < len; i++){if(dp[g[u][i]] > dp[u]+1){dp[g[u][i]] = dp[u]+1;q.push(NODE(dp[g[u][i]],g[u][i]));}}}int ans = 0;for(int i = 0; i < cnt; i++)ans = max(ans,dp[i]);return ans;}int main(){#ifdef SHYfreopen("e:\\1.txt", "r", stdin);#endifint t;scanf("%d%*c", &t);while(t--){scanf("%d %d%*c", &h, &w);for(int i = 0; i < h*w; i++) g[i].clear();memset(f,-1,sizeof(f));for(int i = 0; i < h; i++){for(int j = 0; j < w; j++)map[i][j] = getchar();getchar();}cnt = 0;for(int i = 0; i < h; i++){for(int j = 0; j < w; j++){if(-1 == f[hash(i,j)])dfs(i,j), cnt++;}}int ans = INF;for(int i = 0; i < cnt; i++)ans = min(ans,dijkstra(i));printf("%d\n", ans);}return 0;}

 

【AC CODE(bfs)】130ms

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;#define MAXN 45#define INF 0x3f3f3f3f#define hash(a,b) (a*w+b)#define max(a,b) (a>b?a:b)struct NODE{int dis, num;NODE(int a = 0, int b = 0){dis = a, num = b;}};const int d[4][2]={{-1,0},{1,0},{0,-1},{0,1}};char map[MAXN][MAXN];int h,w, f[MAXN*MAXN], cnt;vector<int> g[MAXN*MAXN];bool vis[MAXN*MAXN];queue<NODE> q;void dfs(int si, int sj){f[hash(si,sj)] = cnt;for(int i = 0; i < 4; i++){int ni = d[i][0]+si, nj = d[i][1]+sj;if(0 <= ni && h > ni && 0 <= nj && w > nj){if(map[ni][nj] == map[si][sj]){if(-1 == f[hash(ni,nj)])dfs(ni,nj);}else if(~f[hash(ni,nj)]){int num = f[hash(ni,nj)];g[cnt].push_back(num);g[num].push_back(cnt);}}}}int bfs(int st){memset(vis,0,sizeof(vis));q.push(NODE(0,st));vis[st] = true;int ans = 0;while(!q.empty()){NODE now = q.front();q.pop();ans = max(ans,now.dis);int len = g[now.num].size();for(int i = 0; i < len; i++){if(!vis[g[now.num][i]]){vis[g[now.num][i]] = true;q.push(NODE(now.dis+1,g[now.num][i]));}}}return ans;}int main(){#ifdef SHYfreopen("e:\\1.txt", "r", stdin);#endifint t;scanf("%d%*c", &t);while(t--){scanf("%d %d%*c", &h, &w);for(int i = 0; i < h*w; i++) g[i].clear();memset(f,-1,sizeof(f));for(int i = 0; i < h; i++){for(int j = 0; j < w; j++)map[i][j] = getchar();getchar();}cnt = 0;for(int i = 0; i < h; i++){for(int j = 0; j < w; j++){if(-1 == f[hash(i,j)])dfs(i,j), cnt++;}}int ans = INF;for(int i = 0; i < cnt; i++)ans = min(ans,bfs(i));printf("%d\n", ans);}return 0;}