2.5 它们其实都是“图”
最短路
AOJ 0189 求图上一点，到所有其他点的距离之和最小
Floyd算法
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 const int inf = 1e8;
5 int d[11][11];
6
7 int main() {
8 int n;
9 while (cin >> n) {
10 if (n == 0) break;
11
12 for (int i = 0; i <= 9; i++)
13 for (int j = 0; j <= 9; j++) d[i][j] = inf;
14 for (int i = 0; i <= 9; i++) d[i][i] = 0;
15
16 int x, y, z, v = 0;
17 while (n--) {
18 cin >> x >> y >> z;
19 v = max(max(x, y), v);
20 d[x][y] = z;
21 d[y][x] = z;
22 }
23
24 for (int k = 0; k <= v; k++) {
25 for (int i = 0; i <= v; i++) {
26 for (int j = 0; j <= v; j++)
27 d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
28 }
29 }
30
31 int sum = inf, number;
32 for (int i = 0; i <= v; i++) {
33 int t = 0;
34 for (int j = 0; j <= v; j++) t += d[i][j];
35 if (sum > t) {
36 sum = t;
37 number = i;
38 }
39 }
40
41 cout << number << " " << sum << "\n";
42 }
43 }
POJ 2139 同上一题，结果乘100除以n-1
1 #include <cstdio>
2 #include <iostream>
3 using namespace std;
4
5 int d[305][305], x[305];
6
7 int main() {
8 int n, m, num;
9 for (int i = 1; i <= 300; i++) fill(d[i], d[i] + 305, 0x7f7f7f);
10 for (int i = 1; i <= 300; i++) d[i][i] = 0;
11 cin >> n >> m;
12 for (int i = 0; i < m; i++) {
13 cin >> num;
14 for (int j = 0; j < num; j++) {
15 cin >> x[j];
16 for (int k = 0; k < j; k++) {
17 d[x[k]][x[j]] = 1;
18 d[x[j]][x[k]] = 1;
19 }
20 }
21 }
22 for (int i = 1; i <= n; i++) {
23 for (int j = 1; j <= n; j++) {
24 for (int k = 1; k <= n; k++)
25 d[j][k] = min(d[j][k], d[j][i] + d[i][k]);
26 }
27 }
28 int sum = (int)1e9;
29 for (int i = 1; i <= n; i++) {
30 int t = 0;
31 for (int j = 1; j <= n; j++) t += d[i][j];
32 sum = min(sum, t);
33 }
34 cout << sum * 100 / (n - 1);
35 }
POJ 3259 判断是否有负环
BF算法
1 #include <cstring>
2 #include <iostream>
3 using namespace std;
4
5 int d[505];
6
7 struct edge {
8 int from, to, cost;
9 };
10
11 edge es[5205];
12
13 int main() {
14 int t, n, m, w, x, y, z;
15 cin >> t;
16 while (t--) {
17 cin >> n >> m >> w;
18 for (int i = 0; i < m; i++) {
19 cin >> x >> y >> z;
20 es[2 * i] = {x, y, z};
21 es[2 * i + 1] = {y, x, z};
22 }
23 for (int i = 2 * m; i < 2 * m + w; i++) {
24 cin >> x >> y >> z;
25 es[i] = {x, y, -z};
26 }
27 memset(d, 0, sizeof(d));
28 int flag = 0;
29 for (int i = 1; i <= n; i++) {
30 for (int j = 0; j < 2 * m + w; j++) {
31 edge e = es[j];
32 if (d[e.to] > d[e.from] + e.cost) {
33 d[e.to] = d[e.from] + e.cost;
34 if (i == n) {
35 cout << "YES\n";
36 flag = 1;
37 break;
38 }
39 }
40 }
41 }
42 if (flag == 0) cout << "NO\n";
43 }
44 }
POJ 3268 求所有点到某一点来回的最短路中的最大值
dij求该点到所有点的最短路，反过来时把所有边反向即可
1 #include <cstring>
2 #include <iostream>
3 using namespace std;
4
5 int d[505];
6
7 struct edge {
8 int from, to, cost;
9 };
10
11 edge es[5205];
12
13 int main() {
14 int t, n, m, w, x, y, z;
15 cin >> t;
16 while (t--) {
17 cin >> n >> m >> w;
18 for (int i = 0; i < m; i++) {
19 cin >> x >> y >> z;
20 es[2 * i] = {x, y, z};
21 es[2 * i + 1] = {y, x, z};
22 }
23 for (int i = 2 * m; i < 2 * m + w; i++) {
24 cin >> x >> y >> z;
25 es[i] = {x, y, -z};
26 }
27 memset(d, 0, sizeof(d));
28 int flag = 0;
29 for (int i = 1; i <= n; i++) {
30 for (int j = 0; j < 2 * m + w; j++) {
31 edge e = es[j];
32 if (d[e.to] > d[e.from] + e.cost) {
33 d[e.to] = d[e.from] + e.cost;
34 if (i == n) {
35 cout << "YES\n";
36 flag = 1;
37 break;
38 }
39 }
40 }
41 }
42 if (flag == 0) cout << "NO\n";
43 }
44 }
AOJ 2249 给出一个连通图，每条边具有距离和修建的成本。要删去一些边，使得从起点1出发到其他的点的最短路径不变 而省下的修建成本尽可能多
可以把修建的成本当做第二维费用 跑dij
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define ll long long
4 #define inc(i, l, r) for (int i = l; i <= r; i++)
5 #define pii pair<int, int>
6 #define fi first
7 #define se second
8 #define pb push_back
9
10 const int maxv = 1e4 + 5;
11 const int inf = 0x3f3f3f3f;
12 int V, d[maxv], c[maxv], m, res;
13 struct edge {
14 int to, dis, cost;
15 bool operator>(const edge& o) const {
16 if (dis == o.dis) return cost > o.cost;
17 return dis > o.dis;
18 }
19 };
20 vector<edge> g[maxv];
21
22 void dij(int s) {
23 priority_queue<edge, vector<edge>, greater<edge> > que;
24 inc(i, 1, V) d[i] = c[i] = inf;
25 d[s] = c[s] = 0;
26 que.push({s, 0, 0});
27 while (!que.empty()) {
28 edge p = que.top();
29 que.pop();
30 int v = p.to;
31 if (d[v] < p.dis || (d[v] == p.dis && c[v] < p.cost)) continue;
32 for (int i = 0; i < g[v].size(); i++) {
33 edge e = g[v][i];
34 if (d[e.to] > d[v] + e.dis ||
35 (d[e.to] == d[v] + e.dis && c[e.to] > e.cost)) {
36 d[e.to] = d[v] + e.dis;
37 c[e.to] = e.cost;
38 que.push({e.to, d[e.to], c[e.to]});
39 }
40 }
41 }
42 }
43
44 int u, v, x, y;
45
46 int main() {
47 while (scanf("%d %d", &V, &m) != EOF && V) {
48 inc(i, 1, V) g[i].clear();
49 inc(i, 1, m) {
50 scanf("%d %d %d %d", &u, &v, &x, &y);
51 g[u].pb({v, x, y});
52 g[v].pb({u, x, y});
53 }
54 dij(1);
55 res = 0;
56 inc(i, 1, V) res += c[i];
57 printf("%d\n", res);
58 }
59 }
AOJ 2200 给出一个图，边有陆路和水路。有个快递员要经过一系列点，他有辆船，走水路时必须用船，走陆路时船得留在出发时的起点，要用船时得回去取。求最短距离
好题目。Floyd分别预处理陆路和水路，dp[i][j]表示当前走到第i个城镇，船停在第j号城镇的最段距离。转移时考虑去取船，走水路，下船走陆路这一过程。特别，可以不取船走陆路到下一个地方
（最开始我dp想当然觉得船都是停在目标路径上的节点，其实不然）
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define ll long long
4 #define inc(i, l, r) for (int i = l; i <= r; i++)
5
6 const int inf = 2e8;
7
8 int n, m, r;
9 int d1[205][205], d2[205][205];
10 int path[1005], dp[1005][205];
11
12 void init(int p[][205]) {
13 inc(i, 1, n) inc(j, 1, n) p[i][j] = inf;
14 inc(i, 1, n) p[i][i] = 0;
15 }
16
17 void floyd(int p[][205]) {
18 inc(k, 1, n) inc(i, 1, n) inc(j, 1, n) p[i][j] =
19 min(p[i][j], p[i][k] + p[k][j]);
20 }
21
22 int x, y, z;
23 char c;
24
25 int main() {
26 while (scanf("%d %d", &n, &m) != EOF && n) {
27 init(d1), init(d2);
28 inc(i, 1, m) {
29 scanf("%d %d %d %c", &x, &y, &z, &c);
30 if (c == ‘L‘)
31 d1[x][y] = d1[y][x] = min(d1[x][y], z);
32 else
33 d2[x][y] = d2[y][x] = min(d2[x][y], z);
34 }
35 floyd(d1), floyd(d2);
36
37 scanf("%d", &r);
38 inc(i, 1, r) scanf("%d", &path[i]);
39
40 inc(i, 1, n) dp[1][i] = inf;
41 dp[1][path[1]] = 0;
42
43 inc(i, 2, r) {
44 inc(j, 1, n) {
45 dp[i][j] = inf;
46 inc(k, 1, n) {
47 dp[i][j] = min(dp[i][j], dp[i - 1][k] + d1[path[i - 1]][k] +
48 d2[k][j] + d1[j][path[i]]);
49 if (k == j)
50 dp[i][j] = min(dp[i][j],
51 dp[i - 1][k] + d1[path[i - 1]][path[i]]);
52 }
53 }
54 }
55
56 int res = inf;
57 inc(i, 1, n) res = min(res, dp[r][i]);
58 printf("%d\n", res);
59 }
60 }
最小生成树
POJ 1258 给出一个图的邻接矩阵，求最小生成树
1 #include <algorithm>
2 #include <cstdio>
3 #include <cstring>
4 #include <iostream>
5 using namespace std;
6
7 int par[105], rankk[105];
8 struct edge {
9 int u, v, cost;
10 } es[10005];
11
12 bool cmp(const edge &x, const edge &y) { return x.cost < y.cost; }
13
14 void init(int x) {
15 for (int i = 1; i <= x; i++) {
16 par[i] = i;
17 rankk[i] = 0;
18 }
19 }
20
21 int find(int x) {
22 if (par[x] == x) return x;
23 return par[x] = find(par[x]);
24 }
25
26 void unite(int x, int y) {
27 x = find(x);
28 y = find(y);
29 if (x == y) return;
30 if (rankk[x] < rankk[y])
31 par[x] = y;
32 else {
33 par[y] = x;
34 if (rankk[x] == rankk[y]) rankk[x]++;
35 }
36 }
37
38 bool same(int x, int y) { return find(x) == find(y); }
39
40 int main() {
41 int n, x, l;
42 while (~scanf("%d", &n)) {
43 init(n);
44 l = 0;
45 memset(es, 0, sizeof(es));
46 for (int i = 1; i <= n; i++) {
47 for (int j = 1; j <= n; j++) {
48 scanf("%d", &x);
49 if (i == j) continue;
50 es[l++] = {i, j, x};
51 }
52 }
53 sort(es, es + l, cmp);
54 int res = 0;
55 for (int i = 0; i < l; i++) {
56 edge e = es[i];
57 if (!same(e.u, e.v)) {
58 unite(e.u, e.v);
59 res += e.cost;
60 }
61 }
62 printf("%d\n", res);
63 }
64 }
POJ 2377 最大生成树
1 #include <algorithm>
2 #include <cstdio>
3 #include <iostream>
4 using namespace std;
5
6 int par[1001], rankk[1001];
7
8 struct edge {
9 int u, v, cost;
10 } es[20001];
11
12 void init(int x) {
13 for (int i = 1; i <= x; i++) {
14 par[i] = i;
15 rankk[i] = 0;
16 }
17 }
18
19 int find(int x) {
20 if (par[x] == x) return x;
21 return par[x] = find(par[x]);
22 }
23
24 void unite(int x, int y) {
25 x = find(x);
26 y = find(y);
27 if (x == y) return;
28 if (rankk[y] > rankk[x]) {
29 par[x] = y;
30 } else {
31 par[y] = x;
32 if (rankk[x] == rankk[y]) rankk[x]++;
33 }
34 }
35
36 bool same(int x, int y) { return find(x) == find(y); }
37
38 bool cmp(edge x, edge y) { return x.cost > y.cost; }
39
40 int main() {
41 int n, m, x, y, z, res = 0, sum = 0;
42 cin >> n >> m;
43 init(n);
44 for (int i = 0; i < m; i++) {
45 scanf("%d%d%d", &x, &y, &z);
46 es[i] = {x, y, z};
47 }
48 sort(es, es + m, cmp);
49 for (int i = 0; i < m; i++) {
50 edge e = es[i];
51 if (!same(e.u, e.v)) {
52 unite(e.u, e.v);
53 sum += e.cost;
54 res++;
55 }
56 }
57 if (res == n - 1)
58 printf("%d", sum);
59 else
60 printf("-1");
61 }
AOJ 2224 给出一个图，要求删去一些边，使得删去后 被这些边划分的所有区域是互相联通的。要使删去的边边长和最小
这题难度在于条件转换。题意等价于图删去一些边后没有环。考虑删边后的图 的边长要尽可能的大，即 是由 若干最大生成树组成的森林
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define ll long long
4 #define inc(i, l, r) for (int i = l; i <= r; i++)
5 #define pii pair<double, int>
6 #define fi first
7 #define se second
8 #define pb push_back
9
10 const double inf = 1e9;
11 const int maxv = 1e5 + 5;
12
13 struct edge {
14 double cost;
15 int u, v;
16 bool operator<(const edge &o) const { return cost > o.cost; }
17 };
18
19 double x[maxv], y[maxv], tot;
20 int n, m, a, b;
21
22 int par[maxv];
23 void init() { inc(i, 1, n) par[i] = i; }
24 int find(int x) {
25 if (x == par[x]) return x;
26 return par[x] = find(par[x]);
27 }
28 void unite(int x, int y) {
29 x = find(x), y = find(y);
30 if (x != y) par[x] = y;
31 }
32 bool same(int x, int y) { return find(x) == find(y); }
33
34 int main() {
35 scanf("%d %d", &n, &m);
36 vector<edge> g(m);
37 inc(i, 1, n) scanf("%lf %lf", &x[i], &y[i]);
38 inc(i, 0, m - 1) {
39 scanf("%d %d", &a, &b);
40 double dis =
41 sqrt((x[a] - x[b]) * (x[a] - x[b]) + (y[a] - y[b]) * (y[a] - y[b]));
42 g[i] = {dis, a, b};
43 tot += dis;
44 }
45 sort(g.begin(), g.end());
46 init();
47 inc(i, 0, m - 1) {
48 int u = g[i].u, v = g[i].v;
49 if (!same(u, v)) {
50 tot -= g[i].cost;
51 unite(u, v);
52 }
53 }
54 printf("%.3f", tot);
55 }
POJ 2395 给定一个图，求它的联通子图（含所有点）最大边的最小值
即最小生成树的最大边
1 #include <algorithm>
2 #include <cstdio>
3 #include <iostream>
4 #include <vector>
5 using namespace std;
6 #define inc(i, l, r) for (int i = l; i <= r; i++)
7
8 const int maxv = 2e3 + 5;
9
10 int par[maxv], n, m, u, v, x;
11
12 struct edge {
13 int u, v, cost;
14 bool operator<(const edge& o) const { return cost < o.cost; }
15 } g[10005];
16
17 void init() { inc(i, 1, n) par[i] = i; }
18 int find(int x) {
19 if (x == par[x]) return x;
20 return par[x] = find(par[x]);
21 }
22 void unite(int x, int y) {
23 x = find(x), y = find(y);
24 if (x != y) par[x] = y;
25 }
26 bool same(int x, int y) { return find(x) == find(y); }
27
28 int main() {
29 scanf("%d %d", &n, &m);
30 init();
31 inc(i, 0, m - 1) {
32 scanf("%d %d %d", &u, &v, &x);
33 g[i] = edge{u, v, x};
34 }
35 sort(g, g + m);
36 int res = 0;
37 inc(i, 0, m - 1) {
38 int u = g[i].u, v = g[i].v;
39 if (!same(u, v)) {
40 unite(u, v);
41 res = max(res, g[i].cost);
42 }
43 }
44 printf("%d\n", res);
45 }
END
原文地址：https://www.cnblogs.com/hs-zlq/p/12233550.html
时间： 2024-10-10 02:04:40