Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
Description
Here is a procedure's pseudocode:
go(int dep, int n, int m)
begin
output the value of dep.
if dep < m and x[a[dep]] + x[b[dep]] != c[dep] then go(dep + 1, n, m)
end
In this code n is an integer. a, b, c and x are 4 arrays of integers. The index of array always starts from 0. Array a and b consist of non-negative integers smaller than n. Array x consists of only 0 and 1. Array c consists of only 0, 1 and 2. The lengths of array a, b and c are mwhile the length of array x is n.
Given the elements of array a, b, and c, when we call the procedure go(0, n , m) what is the maximal possible value does the procedure output?
Input
There are multiple test cases. The first line of input is an integer T (0 < T ≤ 100), indicating the number of test cases. Then T test cases follow. Each case starts with a line of 2 integers n and m (0 < n ≤ 200, 0 < m ≤ 10000). Then m lines of 3 integers follow. The i-th(1 ≤ i ≤m) line of them are ai-1 ,bi-1 and ci-1 (0 ≤ ai-1, bi-1 < n, 0 ≤ ci-1 ≤ 2).
Output
For each test case, output the result in a single line.
Sample Input
3
2 1
0 1 0
2 1
0 0 0
2 2
0 1 0
1 1 2
Sample Output
1
1
2 题目大意：
给定一些方程，x[]数组未知，求前面最多能够有多少方程x[a]+x[b]！=c能够被满足。
其中 c=0,1,2
x[]={0,1}
相当于裸的2sat问题，加上二分
强烈建议阅读 kuangbin大神对2-sat的总结：http://www.cnblogs.com/kuangbin/archive/2012/10/05/2712429.html 总的来说，就是当 a or b 时 连接 a' -> b 与 b' -> a 的边，然后进行强连通判断是否出现 a 与 a' ...在同一个连通分量中，若在则不可能。 建立数a的两个状态，即a与a'，相当于x[a]=1 和 x[a']=0 x[a]+x[b]!=0 => a or b => a'->b 且 a->b'
x[a]+x[b]!=1 => (a and b) or (a' and b') == a or b' 且 a' or b => a'->b' 且 b->a 且 a->b 且 b'->a
x[a]+x[b]!=2 => a' or b' => a->b' 且 a'->b 按上面来建图判断即可
#include<cstdio>
#include<cstring>
int e[],pd[],be[],ne[],all;
int dfn[],low[],instack[],belong[],stack[],stak,curr,num;
int a,b,c,n,m,l,r,mid,flag;
void add(int x,int y,int p){
e[++all]=y;
pd[all]=p;
ne[all]=be[x];
be[x]=all;
}
void tarjan(int x){
instack[x]=;
stack[++stak]=x;
dfn[x]=low[x]=++curr;
for(int j=be[x];j!=;j=ne[j])
if(pd[j]<=mid){
if(!dfn[e[j]]){
tarjan(e[j]);
if(low[x]>low[e[j]]) low[x]=low[e[j]];
}else if(instack[e[j]]&&low[x]>low[e[j]])
low[x]=low[e[j]];
}
if(dfn[x]==low[x]){
int j;
++num;
do{
j=stack[stak--];
instack[j]=;
belong[j]=num;
}while(j!=x);
}
}
int solve(){
curr=stak=num=;
memset(dfn,,sizeof(dfn));
memset(low,,sizeof(low));
memset(instack,,sizeof(instack));
for(int i=;i<*n;i++)
if(!dfn[i]) tarjan(i);
flag=;
for(int i=;i<n;i++)
if(belong[*i]==belong[*i+]){
flag=;
break;
}
return flag;
}
int main()
{
int tt;
scanf("%d",&tt);
while(tt--){
scanf("%d%d",&n,&m);
all=;
memset(e,,sizeof(e));
memset(be,,sizeof(be));
memset(ne,,sizeof(ne));
memset(pd,,sizeof(pd));
for(int i=;i<m;i++){
scanf("%d%d%d",&a,&b,&c);
switch (c){
case : add(*a+,*b,i);
add(*b+,*a,i);
break;
case : add(*a,*b,i);
add(*b+,*a+,i);
add(*b,*a,i);
add(*a+,*b+,i);
break;
case : add(*a,*b+,i);
add(*b,*a+,i);
break;
}
}
l=;r=m-;
while(l<r-){
mid=(l+r)/;
if(solve()) r=mid; else l=mid;
}
mid=r;
if(!solve()) printf("%d\n",r+);
else printf("%d\n",l+);
}
return ;
}