POJ 1679 The Unique MST 次小生成树
题目大意就是问是否有多个权值相同的最小生成树
有两种方法， 一种是枚举删边，然后接着构造最小生成树，但是复杂度比较大
另外一种就比较好了 ， 是求次小生成树的方法， 把生成树上任意两点间的最大边在求最小生成树的同时预处理出来，然后n2的枚举任意两点，如果这两点在最小生成树中不是相邻的，就可以删掉两点间的最大边，换上新边，即他俩之间的直接的边，在邻接矩阵中就是他们的距离。
第一种方法 ，我用克鲁斯卡尔做的，很久前的代码了
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define LOCA
#define MAXN 1005
#define INF 100000000
#define eps 1e-7
using namespace std;
struct wwj
{
int u, v, w;
int equal;
int used;
int del;
}edge[10005];
int parent[10005], num[10005];
int n, m;
bool first;
bool cmp(wwj x, wwj y)
{
return x.w < y.w;
}
void init()
{
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; i++)
{
scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);
edge[i].del = 0; edge[i].equal = 0; edge[i].used = 0;
}
for(int i = 1; i <= m; i++)
{
for(int j = 1; j <= m; j++)
{
if(i == j) continue;
if(edge[j].w == edge[i].w)
edge[i].equal = 1;
}
}
sort(edge + 1, edge + m + 1, cmp);
first = true;
}
int find(int x)
{
if(parent[x] == x)
return x;
int t = find(parent[x]);
parent[x] = t;
return t;
}
void join(int x, int y)
{
int fx = find(x);
int fy = find(y);
if(fx != fy)
{
if(num[fx] > num[fy])
{
parent[fy] = fx;
num[fx] += num[fy];
num[fy] = 1;
}
else
{
parent[fx] = fy;
num[fy] += num[fx];
num[fx] = 1;
}
}
}
int kruskal()
{
int i, sum = 0, cnt = 0, u, v;
for(i = 1; i <= n; i++)
parent[i] = i;
memset(num, 1, sizeof(num));
for(i = 1; i <= m; i++)
{
if(edge[i].del == 1) continue;
u = edge[i].u;
v = edge[i].v;
if(find(u) != find(v))
{
sum += edge[i].w;
cnt++;
join(u, v);
if(first) edge[i].used = 1;
}
if(cnt >= n - 1) break;
}
return sum;
}
void solve()
{
int w1 = kruskal(), w2, i;
first = false;
for(i = 1; i <= m; i++)
{
if(edge[i].used && edge[i].equal)
{
edge[i].del = 1;
w2 = kruskal();
if(w1 == w2)
{
printf("Not Unique!\n");
return;
}
edge[i].del = 0;
}
}
if(i > m) printf("%d\n", w1);
}
int main()
{
#ifdef LOCAL
freopen("d:/data.in","r",stdin);
freopen("d:/data.out","w",stdout);
#endif
int t;
scanf("%d", &t);
while(t--)
{
init();
solve();
}
return 0;
}
第二种方法是用prim做的
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#define MAXN 1005
#define MAXM 100005
#define INF 1000000000
using namespace std;
int mx[105][105];
int d[105][105];
int can[105][105];
int dis[105], used[105], near[105];
int n, m;
void init()
{
for(int i = 1; i <= n; i++)
{
near[i] = 1;
for(int j = 1; j <= n; j++)
{
mx[i][j] = 0;
d[i][j] = INF;
can[i][j] = 0;
}
d[i][i] = 0;
}
memset(used, 0, sizeof(used));
}
int main()
{
int T, x, y, w;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &n, &m);
init();
for(int i = 1; i <= m; i++)
{
scanf("%d%d%d", &x, &y, &w);
d[x][y] = d[y][x] = w;
}
for(int i = 1; i <= n; i++)
dis[i] = d[1][i];
used[1] = true;
near[1] = -1;
int sum = 0;
for(int i = 1; i < n; i++)
{
int mi = INF;
int v = -1;
for(int j = 1; j <= n; j++)
{
if(near[j] != -1 && dis[j] < mi)
{
v = j;
mi = dis[j];
}
}
int pre;
if(v != -1)
{
pre = near[v];
sum += d[v][near[v]];
can[v][near[v]] = can[near[v]][v] = 1;
mx[near[v]][v] = mx[v][near[v]] = d[v][near[v]];
near[v] = -1;
for(int j = 1; j <= n; j++)
{
if(near[j] != -1 && d[v][j] < dis[j])
{
dis[j] = d[v][j];
near[j] = v;
}
}
}
for(int j = 1; j <= n; j++)
{
if(near[j] == -1 && j != v)
{
mx[j][v] = mx[v][j] = max(mx[j][pre], mx[pre][v]);
}
}
}
bool flag = false;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
{
if(i != j && !can[i][j] && d[i][j] < INF)
{
int sum2 = sum - mx[i][j] + d[i][j];
if(sum2 == sum) flag = true;
}
}
if(flag) printf("Not Unique!\n");
else printf("%d\n", sum);
}
return 0;
}