poj3905 Perfect Election c++代码

Perfect Election

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 765		Accepted: 371

Description

In a country (my memory fails to say which), the candidates {1, 2 ..., N} are running in the parliamentary election. An opinion poll asks the question "For any two candidates of your own choice, which election result would make you happy?". The accepted answers are shown in the table below, where the candidates i and j are not necessarily different, i.e. it may happen that i=j. There are M poll answers, some of which may be similar or identical. The problem is to decide whether there can be an election outcome (It may happen that all candidates fail to be elected, or all are elected, or only a part of them are elected. All these are acceptable election outcomes.) that conforms to all M answers. We say that such an election outcome is perfect. The result of the problem is 1 if a perfect election outcome does exist and 0 otherwise.

Input

Each data set corresponds to an instance of the problem and starts with two integral numbers: 1≤N≤1000 and 1≤M≤1000000. The data set continues with M pairs ±i ±j of signed numbers, 1≤i,j≤N. Each pair encodes a poll answer as follows: 

Accepted answers to the poll question	Encoding
I would be happy if at least one from i and j is elected.	+i +j
I would be happy if at least one from i and j is not elected.	-i -j
I would be happy if i is elected or j is not elected or both events happen.	+i -j
I would be happy if i is not elected or j is elected or both events happen.	-i +j

The input data are separated by white spaces, terminate with an end of file, and are correct.

Output

For each data set the program prints the result of the encoded election problem. The result, 1 or 0, is printed on the standard output from the beginning of a line. There must be no empty lines on output.

Sample Input

3 3  +1 +2  -1 +2  -1 -3 
2 3  -1 +2  -1 -2  +1 -2 
2 4  -1 +2  -1 -2  +1 -2  +1 +2 
2 8  +1 +2  +2 +1  +1 -2  +1 -2  -2 +1  -1 +1  -2 -2  +1 -1

Sample Output

1
1
0
1

Hint

For the first data set the result of the problem is 1; there are several perfect election outcomes, e.g. 1 is not elected, 2 is elected, 3 is not elected. The result for the second data set is justified by the perfect election outcome: 1 is not elected, 2 is not elected. The result for the third data set is 0. According to the answers -1 +2 and -1 -2 the candidate 1 must not be elected, whereas the answers +1 -2 and +1 +2 say that candidate 1 must be elected. There is no perfect election outcome. For the fourth data set notice that there are similar or identical poll answers and that some answers mention a single candidate. The result is 1.

题解：

#include<iostream>
#include<cstdio>
#include <algorithm>
#include<cstring>
using namespace std;
const int inf=1<<30;
const int nMax=5015;
const int mMax=10000000;
class edge{
public:
    int v,nex;
};edge e[mMax];
int k,head[nMax];//head[i]是以点i为起点的链表头部


void addedge(int a,int b){//向图中加边的算法，注意加上的是有向边//b为a的后续节点既是a---->b
    e[k].v=b;
    e[k].nex=head[a];
    head[a]=k;k++;
}


int dfn[nMax],low[nMax],sta[nMax],top,atype,belon[nMax],dep;   //atype 强连通分量的个数
bool insta[nMax];


void Tarjan(int u){                 //我的Tarjan模版
    int i,j;
    dfn[u]=low[u]=++dep;
    sta[++top]=u;
    insta[u]=1;
    for(i=head[u];i;i=e[i].nex){
        int v=e[i].v;
        if(!dfn[v]){
            Tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else{
            if(insta[v])low[u]=min(low[u],dfn[v]);
        }
    }
    if(dfn[u]==low[u]){
        atype++;              //强连通分量个数
        do{
            j=sta[top--];
            belon[j]=atype;   //第j个点属于第type个连通块
            insta[j]=0;
        }while(u!=j);
    }
}


int a[nMax];        //每个连通块的出度
int b[nMax];
void init(){
    k=1;
    dep=1;
    top=atype=0;
    memset(insta,0,sizeof(insta)); //是否在栈中
    memset(head,0,sizeof(head));   //静态链表头指针
    memset(low,0,sizeof(low));     //Tarjan的low数组
    memset(dfn,0,sizeof(dfn));     //Tarjan的dfn数组
    memset(belon,0,sizeof(belon)); //记录每个点属于哪一个强连通分量
}


int n,m;


bool judge(){
    for(int i=1;i<=n;i++){
        if(belon[i*2]==belon[i*2+1]){
            return 0;
        }
    }
    return 1;
}
int main(){
    int i,j,a1,a2,c1,c2;
    char str1,str2;
    while(scanf("%d%d",&n,&m)!=EOF){
        init();
        for(i=1;i<=m;i++){
            c1=c2=1;
            scanf("%d%d",&a1,&a2);
            if(a1<0){
                c1=0;
                a1=-a1;
            }
            if(a2<0){
                c2=0;
                a2=-a2;
            }
            addedge(2*a1+c1,2*a2+1-c2);
            addedge(2*a2+c2,2*a1+1-c1);
        }
        for(i=1;i<=2*n+1;i++){
            if(!dfn[i])Tarjan(i);
        }
        if(judge())printf("1");
        else printf("0");
        printf("\n");
    }
    return 0;
}