HDU3231Box Relations

来源：互联网发布：excel表格数据取整编辑：程序博客网时间：2024/05/29 17:41
题目

http://acm.hdu.edu.cn/showproblem.php?pid=3231
第一想法就是拓扑排序，想了一想，发现并不难
代码

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<algorithm>#include<queue>#ifdef L_JUDGE#pragma warning(disable:4996)#endifusing namespace std;#define MAXN 1010#define MAXE 2020int inx[MAXE], iny[MAXE], inz[MAXE];int rex[MAXE], rey[MAXE], rez[MAXE];vector<int> mx[MAXE], my[MAXE], mz[MAXE];int N, R;void init(){    memset(inx, 0, sizeof(inx));    memset(iny, 0, sizeof(iny));    memset(inz, 0, sizeof(inz));    for(int i=2*N;i>=1;i--){        mx[i].clear();        my[i].clear();        mz[i].clear();    }}bool solve(int in[], vector<int> map[], int ans[]){    queue<int> qe;    int maxi=2*N;    for(int i=1;i<=maxi;i++){        if(in[i]==0){            qe.push(i);        }    }    int num=0;    while(!qe.empty()){        int tp=qe.front();        qe.pop();        ans[tp]=num++;        int maxi=map[tp].size()-1;        for(int i=0;i<=maxi;i++){            int temp=map[tp][i];            in[temp]--;            if(0==in[temp]){                qe.push(temp);            }        }    }    if(num==2*N)return true;    else return false;}int main(){    #ifdef L_JUDGE        freopen("in.txt","r",stdin);        freopen("out.txt","w",stdout);    #endif    int casei=1;    while(scanf("%d%d", &N, &R), N+R){        init();        for(int ni=1;ni<=N;ni++){            inx[ni+N]++;            iny[ni+N]++;            inz[ni+N]++;            mx[ni].push_back(ni+N);            my[ni].push_back(ni+N);            mz[ni].push_back(ni+N);        }        char s[5];        int a, b;        for(int ni=1;ni<=R;ni++){            scanf("%s%d%d", s, &a, &b);            switch(s[0]){                case 'I':                    inx[b+N]++;                    inx[a+N]++;                    mx[a].push_back(b+N);                    mx[b].push_back(a+N);                    iny[b+N]++;                    iny[a+N]++;                    my[a].push_back(b+N);                    my[b].push_back(a+N);                    inz[b+N]++;                    inz[a+N]++;                    mz[a].push_back(b+N);                    mz[b].push_back(a+N);                    break;                case 'X':                    inx[b]++;                    mx[a+N].push_back(b);                    break;                case 'Y':                    iny[b]++;                    my[a+N].push_back(b);                    break;                case 'Z':                    inz[b]++;                    mz[a+N].push_back(b);                    break;                default:                    break;            }        }        bool flag=true;        if(!solve(inx, mx, rex))flag=false;        if(!solve(iny, my, rey))flag=false;        if(!solve(inz, mz, rez))flag=false;        if(flag){            printf("Case %d: POSSIBLE\n", casei++);            for(int i=1;i<=N;i++){                printf("%d %d %d %d %d %d\n",                         rex[i], rey[i], rez[i],                         rex[i+N], rey[i+N], rez[i+N] );            }        }else{            printf("Case %d: IMPOSSIBLE\n", casei++);        }        printf("\n");    }    #ifdef L_JUDGE        fclose(stdin);        fclose(stdout);        system("out.txt");    #endif    return 0;}
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