Poj1291 this sentence is false (query set | hash)

From: http://blog.csdn.net/svitter

Before writing:

Recently, I feel that I have done a lot of questions and checked them. It is found that A is right to B, and that games like who is telling the truth, who is telling the truth, are basically a collection query. Make a record to prevent future forgetting.

Question:

N sentences are given in the question. The number starts from 1. Every sentence is in the form of sentence $ num is true/false.

The number of vertices that output the most vertices.

Input and Output Analysis:

Note that the format of scanf cannot be "sentence % d is % s. The specific cause is unknown. You can use % s to constantly read unused variables to eliminate characters in the cache.

Algorithm Data Structure Analysis:

First, I, I + N is used to determine the right and wrong. If there is a paradox, continue to absorb the buffer, but do not perform calculations.

For details, refer to: poj2492 A Bug's Life (and query set)

After finding the corresponding set, count the number of num [getroot (I)] and add them together.

Problem:

At first, I also counted getroot (I + n). Later, I referred to the code of the great god. Considering that I + N is an opposite situation of I, there is no need to calculate it.

The confrontation with I must be included in the case except I, so there is no need to make statistics.

In the process of traversing the num array, take the larger group of num [getroot (I)] and num [getroot (I + n. That is, take the I or ~ I. Because there is no paradox, I and ~ I must have been established.

AC code:

//author: svtter//#include <iostream>#include <stdio.h>#include <string.h>#include <vector>#include <map>#include <algorithm>#include <queue>#include <cmath>#define INF 0xffffffusing namespace std;int root[2010];int n;int num[2010];int vis[2010];void init(){    for(int i = 0; i <= n*2; i++)        root[i] = i;}int getRoot(int i){    if(i == root[i])        return i;    return root[i] = getRoot(root[i]);}int a, b;void Merge(int i, int j){    a = getRoot(i);    b = getRoot(j);    if(a == b)        return;    root[a] = b;}bool jud(int i, int j){    return getRoot(i) == getRoot(j);}void solve(){    int i, ans;     ans = 0;    memset(num, 0, sizeof(num));    memset(vis, 0, sizeof(vis));    for(i = 1; i <= n; i++)    {        num[getRoot(i)] ++;           // num[getRoot(i+n)]++;    }        for(i = 1; i <= n; i++)    {        if(vis[getRoot(i)] || vis[getRoot(i+n)])            continue;        ans += max(num[getRoot(i)], num[getRoot(i+n)]);        vis[getRoot(i)] = vis[getRoot(i+n)] = 1;    }    printf("%d\n", ans);}int main(){    char ju[10];    int i, t;    bool paradox;    while(~scanf("%d", &n))    {        if(n == 0)            break;        paradox = false;        //input        init();        for(i = 1; i <= n; ++i)        {            scanf("%s", ju);            scanf("%d", &t);            scanf("%s", ju);            scanf("%s", ju);            if(paradox)                continue;            if(ju[0] == 'f')            {                if(jud(i, t) || jud(i+n, t+n))                    paradox = 1;                else                {                    Merge(i, t+n);                    Merge(t, i+n);                }            }            else            {                if(jud(i, t+n) || jud(t, i+n))                    paradox = 1;                else                {                    Merge(i, t);                    Merge(i+n, t+n);                }            }        }        if(paradox)            printf("Inconsistent\n");        else            solve();    }    return 0;}