存在邊權(quán)為負的情況下，無法求最小生成樹
裸題：1140. 最短網(wǎng)絡(luò)
1140. 最短網(wǎng)絡(luò) - AcWing題庫
套個prim的板子即可
#include <iostream>
#include <cstring>
using namespace std;
const int N = 110, INF = 0x3f3f3f3f;
int g[N][N];
int dis[N]; bool st[N];
int n;
int prim()
{
memset(dis, 0x3f, sizeof(dis));
int res = 0;
for (int i = 0; i < n; ++ i )
{
int x = -1;
for (int j = 1; j <= n; ++ j )
if (!st[j] && (x == -1 || dis[x] > dis[j])) x = j;
st[x] = true;
if (i && dis[x] == INF) return INF;
if (i) res += dis[x];
for (int y = 1; y <= n; ++ y )
dis[y] = min(dis[y], g[x][y]);
}
return res;
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; ++ i )
for (int j = 1; j <= n; ++ j )
scanf("%d", &g[i][j]);
printf("%d\n", prim());
return 0;
}
裸題：1141. 局域網(wǎng)
1141. 局域網(wǎng) - AcWing題庫
裸題，稀疏圖，套個kruskal的板子就行
需要注意的是：題目給定的圖可能存在多個連通塊，若使用prim算法，需要對每個連通塊求最小生成樹，但是使用kruskal能直接求出所有連通塊的最小生成樹
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 110, M = 210;
struct Edge
{
int x, y, w;
bool operator<(const Edge& e) const
{
return w < e.w;
}
}edges[M];
int p[N];
int n, m;
int find(int x)
{
if (x != p[x]) p[x] = find(p[x]);
return p[x];
}
int kruskal()
{
sort(edges, edges + m);
for (int i = 1; i <= n; ++ i ) p[i] = i;
int cnt = 0, res = 0;
for (int i = 0; i < m; ++ i )
{
auto t = edges[i];
int x = edges[i].x, y = edges[i].y, w = edges[i].w;
x = find(x), y = find(y);
if (x != y)
{
cnt ++ ;
res += w;
p[x] = y;
}
}
return res;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 0; i < m; ++ i )
scanf("%d%d%d", &edges[i].x, &edges[i].y, &edges[i].w);
int sum = 0;
for (int i = 0; i < m; ++ i ) sum += edges[i].w;
printf("%d\n", sum - kruskal());
return 0;
}
裸題：1142. 繁忙的都市
1142. 繁忙的都市 - AcWing題庫
依然是套kruskal的板子
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 310 ,M = 8010;
struct Edge
{
int x, y, w;
bool operator<(const Edge& e) const
{
return w < e.w;
}
}edges[M];
int n, m;
int p[N];
int find(int x)
{
if (x != p[x]) p[x] = find(p[x]);
return p[x];
}
int kruskal()
{
sort(edges, edges + m);
int res = 0;
for (int i = 1; i <= n; ++ i ) p[i] = i;
for (int i = 0; i < m; ++ i )
{
auto t = edges[i];
int x = t.x, y = t.y, w = t.w;
x = find(x), y = find(y);
if (x != y)
{
res = max(res, w);
p[x] = y;
}
}
return res;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 0; i < m; ++ i )
scanf("%d%d%d", &edges[i].x, &edges[i].y, &edges[i].w);
printf("%d %d\n", n - 1, kruskal());
return 0;
}
裸題：1143. 聯(lián)絡(luò)員
1143. 聯(lián)絡(luò)員 - AcWing題庫
添加所有必選的邊，維護并查集，然后再對非必選的邊做kruskal
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 2010, M = 10010;
struct Edge
{
int x, y, w;
bool operator<(const Edge& e) const
{
return w < e.w;
}
}edges[M];
int n, m, cnt;
int p[N];
int find(int x)
{
if (x != p[x]) p[x] = find(p[x]);
return p[x];
}
int kruskal()
{
int res = 0;
for (int i = 0; i < cnt; ++ i )
{
auto t = edges[i];
int x = t.x, y = t.y, w = t.w;
x = find(x), y = find(y);
if (x != y)
{
res += w;
p[x] = y;
}
}
return res;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++ i ) p[i] = i;
int t, x, y, d;
int res = 0;
while ( m -- )
{
scanf("%d%d%d%d", &t, &x, &y, &d);
if (t == 1)
{
x = find(x), y = find(y);
if (x != y) p[x] = y;
res += d;
}
else
{
edges[cnt].x = x, edges[cnt].y = y, edges[cnt].w = d;
cnt ++ ;
}
}
sort(edges, edges + cnt);
res += kruskal();
printf("%d\n", res);
return 0;
}
有些麻煩的裸題：1144. 連接格點
1144. 連接格點 - AcWing題庫
點陣為圖中的點，將二維坐標(biāo)轉(zhuǎn)換成一維，作為點的編號
添加已有連線后，做kruskal即可
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
struct Edge
{
int x, y, w;
bool operator<(const Edge& e) const
{
return w < e.w;
}
}edges[2 * N * N];
int n, m, cnt = 1;
int p[N * N];
int g[N][N]; // 二維到一維
int dx[3] = { 0, 1, 0 }, dy[3] = { 0, 0, 1 };
int find(int x)
{
if (x != p[x]) p[x] = find(p[x]);
return p[x];
}
int kruskal()
{
int res = 0;
for (int i = 0; i < cnt; ++ i )
{
auto t = edges[i];
int x = t.x, y = t.y, w = t.w;
x = find(x), y = find(y);
if (x != y)
{
res += w;
p[x] = y;
}
}
return res;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++ i )
for (int j = 1; j <= m; ++ j )
g[i][j] = cnt ++ ;
for (int i = 1; i < cnt; ++ i ) p[i] = i;
int x1, x2, y1, y2;
while (~scanf("%d%d%d%d", &x1, &y1, &x2, &y2))
{
int x = g[x1][y1], y = g[x2][y2];
x = find(x), y = find(y);
if (x != y) p[x] = y;
}
cnt = 0;
for (int i = 1; i <= n; ++ i )
for (int j = 1; j <= m; ++ j )
for (int k = 1; k <= 2; ++ k )
{
int a = i + dx[k], b = j + dy[k];
if (a >= 1 && a <= n && b >= 1 && b <= m)
{
int x = g[i][j], y = g[a][b];
edges[cnt ++ ] = { x, y, k };
}
}
sort(edges, edges + cnt);
printf("%d\n", kruskal());
return 0;
}
debug：n * m的矩陣中，相鄰兩點之間存在一條邊，那么矩陣中的邊數(shù)應(yīng)該為m(n-1) + n(m-1)，大概就是2 * n * n，數(shù)組開小了導(dǎo)致SF
盡量不要在for循環(huán)中定義除了循環(huán)變量之外的變量文章來源：http://www.zghlxwxcb.cn/news/detail-628344.html
需要注意的是，200萬條邊進行排序會消耗很多時間，由于邊的權(quán)值只有1和2，所以可以先添加權(quán)值為1的邊，再添加權(quán)值為2的邊文章來源地址http://www.zghlxwxcb.cn/news/detail-628344.html
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