称号：
Cat vs. Dog |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 219 Accepted Submission(s): 86 |
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Problem Description The latest reality show has hit the TV: ``Cat vs. Dog‘‘. In this show, a bunch of cats and dogs compete for the very prestigious Best Pet Ever title. In each episode, the cats and dogs get to show themselves off, after which the viewers vote on which pets should stay and which should be forced to leave the show. Each viewer gets to cast a vote on two things: one pet which should be kept on the show, and one pet which should be thrown out. Also, based on the universal fact that everyone is either a cat lover (i.e. a dog hater) or a dog lover (i.e. a cat hater), it has been decided that each vote must name exactly one cat and exactly one dog. Ingenious as they are, the producers have decided to use an advancement procedure which guarantees that as many viewers as possible will continue watching the show: the pets that get to stay will be chosen so as to maximize the number of viewers who get both their opinions satisfied. Write a program to calculate this maximum number of viewers. |
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Input On the first line one positive number: the number of testcases, at most 100. After that per testcase: * One line with three integers c, d, v (1 ≤ c, d ≤ 100 and 0 ≤ v ≤ 500): the number of cats, dogs, and voters. |
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Output Per testcase: * One line with the maximum possible number of satisfied voters for the show. |
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Sample Input 2 1 1 2 C1 D1 D1 C1 1 2 4 C1 D1 C1 D1 C1 D2 D2 C1 |
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Sample Output 1 3 |
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Source NWERC 2008 |
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Recommend lcy |
题目大意：
有v个观众，分别投出给自己喜欢的动物和讨厌的动物。
假设一个观众喜欢的动物和另外一个观众喜欢的动物发生冲突，则让一个观众离开，问最后可以留下几个观众。
、
例子分析：
输入分析：
2 1 1 2 C1 D1//第一个观众喜欢的是Cat1 讨厌的是Dog1 D1 C1 1 2 4 C1 D1 C1 D1 C1 D2 D2 C1
输出分析：
在第二个例子中。输出结果3是怎么得到的呢？我们把喜欢猫的放在一边喜欢狗的放在一边。然后依据冲突的产生情况建边，这时候我们就能得到下图：
这时候最大匹配就是1----4, 2----4 ， 3-----4 这三条边中的一条。上图选择了1----4这条边来举例。假设还想不明确的同学，在这里我们复习一下相关概念
匹配：是一个边的集合，随意两条边不存在公共点。
最大匹配：变数最多的匹配。
题目分析：
把喜欢猫的放在一边，把喜欢狗的放在一边，若发生了冲突。则建一条边。则问题转化成二分图的最大独立集问题。
须要注意一下的是这道题中的建边规则。
这道题使用邻接矩阵实现的。耗时好像是390ms。用邻接表可能会快一些。
代码例如以下：
/*
* g.cpp
*
* Created on: 2015年3月14日
* Author: Administrator
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 501;
int map[maxn][maxn];
bool useif[maxn];
int link[maxn];
int n;
bool can(int t){
int i;
for(i = 1 ; i <= n ; ++i){
if(useif[i] == false && map[t][i] == true){
useif[i] = true;
if(link[i] == -1 || can(link[i])){
link[i] = t;
return true;
}
}
}
return false;
}
int max_match(){
int num = 0;
int i;
for(i = 1 ; i <= n ; ++i){
memset(useif,false,sizeof(useif));
if(can(i) == true){
num++;
}
}
return num;
}
int main(){
string loves[maxn];
string hates[maxn];
int t;
scanf("%d",&t);
while(t--){
memset(map,false,sizeof(map));
memset(link,-1,sizeof(link));
int cats,dogs;
scanf("%d%d%d",&cats,&dogs,&n);
int i;
for(i = 1 ; i <= n ; ++i){
cin >> loves[i] >> hates[i];
}
/**
* 这道题与其它题不同的地方就在于建边部分.
* 所以其它部分不再加凝视,在这里仅仅为建边过程加凝视
*/
int j;
for(i = 1 ; i <= n ; ++i){//遍历每个观众的喜欢的动物
for(j = 1 ; j <= n ; ++j){//遍历每个观众讨厌的动物
//假设第i个观众喜欢的动物==第j个观众讨厌的动物 || 第i个观众讨厌的动物 == 第j个观众喜欢的动物。
证明i和j之间存在冲突
if(loves[i] == hates[j] || hates[i] == loves[j]){
//为i和j建边
map[i][j] = true;
map[j][i] = true;
}
}
}
printf("%d\n",n - max_match()/2);
}
return 0;
}
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