AOJ0531(imos-累积和法)

来源：互联网发布：淘宝助理5天猫用不了编辑：程序博客网时间：2024/05/19 15:21

来自《挑战程序设计竞赛》

但是找到的题解和书上给的模板很不一样……

题解参考了http://www.hankcs.com/program/algorithm/aoj-0531-paint-color.html

给出题目的链接http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0531

突然发现，写博客好麻烦，尤其是要做表格什么的╮(╯_╰)╭

AC代码:

#include<algorithm>#include<cctype>#include<cmath>#include<cstdio>#include<cstdlib>#include<cstring>#include<iomanip>#include<iostream>#include<map>#include<queue>#include<string>#include<set>#include<vector>#include<cmath>#include<bitset>#include<stack>#include<sstream>using namespace std;#define INF 0x7fffffffconst int maxn=1005;int dx[]={1,-1,0,0};int dy[]={0,0,1,-1};int W,H,N;int X1[maxn],X2[maxn],Y1[maxn],Y2[maxn];//填充用int fld[2*maxn][2*maxn];//对x1和x2进行坐标离散化//返回离散化之后的宽度int compress(int *x1,int *x2,int w){    vector<int> xs;    for(int i=0;i<N;i++){        int tx1=x1[i];        int tx2=x2[i];        if(1<=tx1&&tx1<w) xs.push_back(tx1);        if(1<=tx2&&tx2<w) xs.push_back(tx2);    }    xs.push_back(0);    xs.push_back(w);    //排序去重    sort(xs.begin(),xs.end());    xs.erase(unique(xs.begin(),xs.end()),xs.end());    for(int i=0;i<N;i++){        x1[i]=find(xs.begin(),xs.end(),x1[i])-xs.begin();        x2[i]=find(xs.begin(),xs.end(),x2[i])-xs.begin();    }    return xs.size()-1;}//宽度优先搜索//求区域的个数int bfs(){    int ans=0;    for(int y=0;y<H;y++){        for(int x=0;x<W;x++){            if(fld[y][x]) continue;            ans++;            queue<pair<int,int> > que;            que.push(make_pair(x,y));            while(!que.empty()){                int sx=que.front().first;                int sy=que.front().second;                que.pop();                for(int i=0;i<4;i++){                    int tx=sx+dx[i];                    int ty=sy+dy[i];                    if(tx<0||tx>W||ty<0||ty>H) continue;                    if(fld[ty][tx]>0) continue;                    que.push(make_pair(tx,ty));                    fld[ty][tx]=1;                }            }        }    }    return ans;}void solve(){    memset(fld,0,sizeof(fld));    //坐标离散化    W=compress(X1,X2,W);    H=compress(Y1,Y2,H);    //imos法    for(int i=0;i<N;i++){        fld[Y1[i]][X1[i]]++;        fld[Y1[i]][X2[i]]--;        fld[Y2[i]][X1[i]]--;        fld[Y2[i]][X2[i]]++;    }    //横向累积    for(int i=0;i<H;i++){        for(int j=1;j<W;j++){            fld[i][j]+=fld[i][j-1];        }    }    //纵向累积    for(int i=1;i<H;i++){        for(int j=0;j<W;j++){            fld[i][j]+=fld[i-1][j];        }    }    //累积完后，fld非0部分表示有挡板    cout<<bfs()<<endl;}int main(){    while(scanf("%d%d",&W,&H)==2){        if(W==0&&H==0) break;        scanf("%d",&N);        for(int i=0;i<N;i++){            scanf("%d%d%d%d",&X1[i],&Y1[i],&X2[i],&Y2[i]);        }        solve();    }    return 0;}

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