POJ 2377 Bad Cowtractors(最大生成树)
2015-11-08 11:07
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题目地址:点击打开链接
思路:就是求最大生成树,即把n个点都连起来且不会出现回路,并且花费最大,刚开是思维定式了,把边赋初值为一个最大值,结果调了半天,比赛完了5分钟就A了,把初始边都赋值为0,每次就最大边加入,还有就是图类问题要注意多重边的可能,这道题要选最大的边
AC代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>
typedef long long ll;
using namespace std;
const int maxn = 1e3+10;
int map1[maxn][maxn];
int lowdist[maxn];
int visit[maxn];
int n,m,sum;
bool flag;
void Kruskal()
{
memset(visit,0,sizeof(visit));
int i,j,k;
for(i=1; i<=n; i++)
{
lowdist[i] = map1[1][i];
}
visit[1] = 1;
for(i=1; i<n; i++)//要并入n个点所以要循环n次
{
int max2 = 0;
k = 0;
for(j=1; j<=n; j++)
{
if(!visit[j] && lowdist[j] > max2)
{
max2 = lowdist[j];
k = j;
}
}
if(k == 0)
{
flag = false;
return;
}
visit[k] = 1;
sum += lowdist[k];
for(j=1; j<=n; j++)
{
if(!visit[j] && map1[k][j] > lowdist[j])
{
lowdist[j] = map1[k][j];
}
}
}
}
int main()
{
int i;
while(scanf("%d%d",&n,&m) != EOF)
{
memset(map1,0,sizeof(map1));
int a,b,c;
for(i=1; i<=m; i++)
{
scanf("%d%d%d",&a,&b,&c);
if(c > map1[a][b])
{
map1[a][b] = map1[b][a] = c;
}
}
flag = true;
sum = 0;
Kruskal();
if(flag)
{
printf("%d\n",sum);
}
else
{
printf("-1\n");
}
}
return 0;
}
错误代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>
typedef long long ll;
using namespace std;
const int zui = 1e6+10;
const int maxn = 1e3+10;
int map1[maxn][maxn];
int lowdist[maxn];
int visit[maxn];
int n,m,sum;
bool flag;
void Kruskal()
{
memset(visit,0,sizeof(visit));
int i,j,k;
for(i=1; i<=n; i++)
{
lowdist[i] = map1[1][i];
}
visit[1] = 1;
for(i=1; i<n; i++)//要并入n个点所以要循环n次
{
int max2 = 0;
k = 0;
for(j=1; j<=n; j++)
{
if(!visit[j] && lowdist[j] >max2 && lowdist[j] <= 100000)
{
max2 = lowdist[j];
k = j;
}
}
if(k == 0)
{
flag = false;
return;
}
visit[k] = 1;
sum += lowdist[k];
for(j=1; j<=n; j++)
{
if(!visit[j] && map1[k][j] > lowdist[j] && map1[k][j] <=100000)
{
lowdist[j] = map1[k][j];
}
}
}
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m) != EOF)
{
for(i=1; i<=n; i++)
{
for(j=i+1; j<=n; j++)
{
map1[i][j] = map1[j][i] = zui;
}
map1[i][i] = 0;
}
int a,b,c;
for(i=1; i<=m; i++)
{
scanf("%d%d%d",&a,&b,&c);
map1[a][b] = map1[b][a] = c;
}
flag = true;
sum = 0;
Kruskal();
if(flag)
{
printf("%d\n",sum);
}
else
{
printf("-1\n");
}
}
return 0;
}
没想好,就知道无脑敲代码
思路:就是求最大生成树,即把n个点都连起来且不会出现回路,并且花费最大,刚开是思维定式了,把边赋初值为一个最大值,结果调了半天,比赛完了5分钟就A了,把初始边都赋值为0,每次就最大边加入,还有就是图类问题要注意多重边的可能,这道题要选最大的边
AC代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>
typedef long long ll;
using namespace std;
const int maxn = 1e3+10;
int map1[maxn][maxn];
int lowdist[maxn];
int visit[maxn];
int n,m,sum;
bool flag;
void Kruskal()
{
memset(visit,0,sizeof(visit));
int i,j,k;
for(i=1; i<=n; i++)
{
lowdist[i] = map1[1][i];
}
visit[1] = 1;
for(i=1; i<n; i++)//要并入n个点所以要循环n次
{
int max2 = 0;
k = 0;
for(j=1; j<=n; j++)
{
if(!visit[j] && lowdist[j] > max2)
{
max2 = lowdist[j];
k = j;
}
}
if(k == 0)
{
flag = false;
return;
}
visit[k] = 1;
sum += lowdist[k];
for(j=1; j<=n; j++)
{
if(!visit[j] && map1[k][j] > lowdist[j])
{
lowdist[j] = map1[k][j];
}
}
}
}
int main()
{
int i;
while(scanf("%d%d",&n,&m) != EOF)
{
memset(map1,0,sizeof(map1));
int a,b,c;
for(i=1; i<=m; i++)
{
scanf("%d%d%d",&a,&b,&c);
if(c > map1[a][b])
{
map1[a][b] = map1[b][a] = c;
}
}
flag = true;
sum = 0;
Kruskal();
if(flag)
{
printf("%d\n",sum);
}
else
{
printf("-1\n");
}
}
return 0;
}
错误代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>
typedef long long ll;
using namespace std;
const int zui = 1e6+10;
const int maxn = 1e3+10;
int map1[maxn][maxn];
int lowdist[maxn];
int visit[maxn];
int n,m,sum;
bool flag;
void Kruskal()
{
memset(visit,0,sizeof(visit));
int i,j,k;
for(i=1; i<=n; i++)
{
lowdist[i] = map1[1][i];
}
visit[1] = 1;
for(i=1; i<n; i++)//要并入n个点所以要循环n次
{
int max2 = 0;
k = 0;
for(j=1; j<=n; j++)
{
if(!visit[j] && lowdist[j] >max2 && lowdist[j] <= 100000)
{
max2 = lowdist[j];
k = j;
}
}
if(k == 0)
{
flag = false;
return;
}
visit[k] = 1;
sum += lowdist[k];
for(j=1; j<=n; j++)
{
if(!visit[j] && map1[k][j] > lowdist[j] && map1[k][j] <=100000)
{
lowdist[j] = map1[k][j];
}
}
}
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m) != EOF)
{
for(i=1; i<=n; i++)
{
for(j=i+1; j<=n; j++)
{
map1[i][j] = map1[j][i] = zui;
}
map1[i][i] = 0;
}
int a,b,c;
for(i=1; i<=m; i++)
{
scanf("%d%d%d",&a,&b,&c);
map1[a][b] = map1[b][a] = c;
}
flag = true;
sum = 0;
Kruskal();
if(flag)
{
printf("%d\n",sum);
}
else
{
printf("-1\n");
}
}
return 0;
}
没想好,就知道无脑敲代码
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