HDU 5925 Coconuts(二维离散化权值统计 经典)

题目链接

这是2016 ccpc 东北的银牌题，很经典.
Coconuts

分析

直接二维离散化,然后记录下各压缩了多少行和列，将其权值相乘便是离散化的图里的权重. dfs or bfs 统计一下就行了.

//Problem : 5925 ( Coconuts )     Judge Status : Accepted//RunId : 22331126    Language : G++    Author : zouzhitao//Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta#include<bits/stdc++.h>#define pb push_back#define mp make_pair#define PI acos(-1)#define fi first#define se second#define INF 0x3f3f3f3f#define INF64 0x3f3f3f3f3f3f3f3f#define random(a,b) ((a)+rand()%((b)-(a)+1))#define ms(x,v) memset((x),(v),sizeof(x))#define eps 1e-8using namespace std;typedef unsigned long long ULL;typedef long long LL;typedef long double DB;typedef pair<int,int> Pair;const int maxn = 400+10;std::map<int, int> idx_x,idx_y;LL vx[maxn],vy[maxn];int a[maxn][maxn];std::vector<LL> ans;int x[maxn],y[maxn];Pair bad[maxn];int lishan(int * arr,int len, LL *val,std::map<int, int>& m ){    m.clear();    sort(arr,arr+len);    len = unique(arr,arr+len)-arr;    int t =1;    val[t] = 1;    m[arr[0]] = t++;    for(int i=1 ; i<len ; ++i){        if(arr[i]!=arr[i-1]+1)val[t++] = arr[i] - arr[i-1]-1;//补空白        val[t] =1;m[arr[i]] = t++;    }    return t;}int dx[] = {1,0,0,-1};int dy[] = {0,1,-1,0};int vis[maxn][maxn];LL bfs(int i,int j) {    queue<Pair> Q;    Q.push(mp(i,j));    vis[i][j] =1;    LL cnt =0;    while (!Q.empty()) {        Pair p = Q.front();Q.pop();        cnt += vx[p.fi] * vy[p.se];        for(int i=0 ; i<4 ; ++i){            int ni = p.fi + dx[i];            int nj = p.se+dy[i];            if(!a[ni][nj] && !vis[ni][nj]){                vis[ni][nj] =1;                Q.push(mp(ni,nj));            }        }    }    return cnt;}int main(){    // ios_base::sync_with_stdio(0);    // cin.tie(0);    // cout.tie(0);    int T;    cin>>T;    for(int kase =1 ; kase <=T ; ++kase){        int n,m,k;        cin>>n>>m>>k;        for(int i=1 ; i<=k ; ++i){            scanf("%d%d",&bad[i].fi, &bad[i].se );            x[i] = bad[i].fi;y[i] = bad[i].se;        }        x[0] = 1;x[k+1] = n;        y[0] = 1;y[k+1] = m;        n = lishan(x,k+2,vx,idx_x);        m = lishan(y,k+2,vy,idx_y);        ms(a,0);        for(int i=1;  i<=k ; ++i)a[idx_x[bad[i].fi]][idx_y[bad[i].se]] = 1;        for(int i=0 ; i<=m ; ++i)a[0][i] = a[n][i] =1;        for(int i=0 ; i<=n ; ++i)a[i][0] = a[i][m] =1;        // std::cout << "val" << '\n';        // for(int i=0 ; i<=n; ++i){        //     for(int j=0 ; j<=m ; ++j)std::cout << vx[i]*vy[j] << ' ';std::cout << '\n';        // }        // std::cout << "mat" << '\n';        // for(int i=0 ; i<=n ; ++i){        //     for(int j=0 ; j<=m ; ++j)std::cout << a[i][j] << ' ';std::cout << '\n';        // }        // std::cout << "vx vy" << '\n';        // for(int i=0 ; i<=n ; ++i)std::cout << vx[i] << ' ';std::cout  << '\n';        // for(int i=0 ; i<=m ; ++i)std::cout << vy[i] << ' ';std::cout  << '\n';        ms(vis,0);        ans.clear();        for(int i=1 ; i<n ; ++i){            for(int j=1 ; j<m ; ++j){                if(!a[i][j] && !vis[i][j])ans.pb(bfs(i,j));            }        }        printf("Case #%d:\n",kase );        sort(ans.begin(),ans.end());        printf("%d\n",ans.size() );        for(unsigned int i=0 ; i<ans.size() ; ++i){            printf("%lld%c",ans[i],i==ans.size()-1?'\n':' ' );        }    }    return 0;}