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Catalog
Problem：
原题目描述在最下面。
第一题：
给定一张n(3000)个点的图，q次询问，每次询问增大一条边的权值后最小生成树的值是多少，求答案期望。
第二题：
给定一张n(1000)个点的图，某两个点间的边可能不存在，问各种情况最小生成树的最大值。
Solution：
第一题用的(kruskal)，第二题用的(prim)。貌似(prim)算法效率更高，毕竟都是稠密图。
先求MST，跑一边树形(dp)，(dp)数组记录任意两个路径间不在MST上的最小边权值。
然后就枚举边求解咯。
AC_Code：
hdu4126
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <ctime>
#include <cmath>
#include <iostream>
#define lson(x) cw[x].l
#define rson(x) cw[x].r
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MXN = 3e3 + 5;
const int MXE = 1e7 + 5;
const int INF = 0x3f3f3f3f;
int n, m, tot;
vector<int> son[MXN];
struct lp{
int u, v;
int w;
}edge[MXE];
int fa[MXN], rnk[MXN];
int dis1[MXN][MXN], is[MXN][MXN], ar[MXN][MXN];
void init() {
memset(dis1, 0x3f, sizeof(dis1));
memset(is, 0, sizeof(is));
memset(ar, 0x3f, sizeof(ar));
for(int i = 1; i <= n; ++i) {
fa[i] = i;
rnk[i] = 1;
son[i].clear();
ar[i][i] = dis1[i][i] = 0;
}
}
bool cmp(const lp &a, const lp &b) {
return a.w < b.w;
}
int Fi(int x) {
return fa[x] == x? x: fa[x] = Fi(fa[x]);
}
LL kruskal(int k) {
int CNT = 0;
LL ans = 0;
for(int i = 0; i < k; ++i) {
int x = Fi(edge[i].u), y = Fi(edge[i].v);
if(x == y) continue;
if(rnk[x] > rnk[y]) swap(x, y);
fa[x] = y;
rnk[y] += rnk[x];
++ CNT;
ans += edge[i].w;
son[edge[i].u].push_back(edge[i].v);
son[edge[i].v].push_back(edge[i].u);
is[edge[i].u][edge[i].v] = is[edge[i].v][edge[i].u] = 1;
if(CNT == n-1) break;
}
return ans;
}
int dfs(int u,int ba,int r) {
int len = son[u].size(), mi = INF;
for(int i = 0; i < len; ++i) {
int v = son[u][i];
if(v == ba) continue;
int tmp = dfs(v, u, r);
mi = min(mi, tmp);
dis1[u][v] = dis1[v][u] = min(dis1[v][u], tmp);
}
if(r != ba) mi = min(mi, ar[r][u]);
return mi;
}
void solve() {
init();
for(int i = 0; i < m; ++i) {
scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);
edge[i].u ++; edge[i].v ++;
ar[edge[i].u][edge[i].v] = ar[edge[i].v][edge[i].u] = edge[i].w;
}
sort(edge, edge + m, cmp);
int ans = kruskal(m), q;
for(int i = 1; i <= n; ++i) dfs(i, -1, i);
LL sum = 0, tmp;
scanf("%d", &q);
for(int i = 0, u, v, w; i < q; ++i) {
scanf("%d%d%d", &u, &v, &w);
u++; v++;
if(is[u][v]) {
tmp = ans - ar[u][v] + min(dis1[u][v], w);
}else {
tmp = ans;
}
sum += tmp;
}
printf("%.4f
", sum*1.0/q);
}
int main(int argc, char const *argv[]){
while(~scanf("%d%d", &n, &m)&&(n + m)) solve();
return 0;
}
hdu4756
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <ctime>
#include <cmath>
#include <iostream>
#define lson(x) cw[x].l
#define rson(x) cw[x].r
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MXN = 1e3 + 5;
const int MXE = 1e7 + 5;
const int INF = 0x3f3f3f3f;
int n;
double m;
vector<int> son[MXN];
int is[MXN][MXN];
double ar[MXN][MXN], dis1[MXN][MXN], dis[MXN];
int pre[MXN], vis[MXN];
LL x[MXN], y[MXN];
void init() {
memset(is, 0, sizeof(is));
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= n; ++j) ar[i][j] = dis1[i][j] = INF;
son[i].clear();
ar[i][i] = dis1[i][i] = 0;
}
}
double prim() {
double sum = 0;
for(int i = 1; i <= n; ++i) {
dis[i] = ar[1][i];
vis[i] = 1;
pre[i] = 1;
}
vis[1] = -1;
for(int J = 1; J < n; ++J) {
int p = -1;
double mm = INF;
for(int i = 1; i <= n; ++i) {
if(vis[i] != -1 && dis[i] < mm) {
p = i; mm = dis[i];
}
}
if(p == -1) break;
sum += mm;
son[p].push_back(vis[p]);
son[vis[p]].push_back(p);
is[p][vis[p]] = is[vis[p]][p] = 1;
vis[p] = -1;
for(int i = 1; i <= n; ++i) {
if(ar[p][i] < dis[i] && vis[i] != -1) {
dis[i] = ar[p][i];
vis[i] = p;
}
}
}
return sum;
}
double dfs(int u,int ba,int r) {
int len = son[u].size();
double mi = INF;
for(int i = 0; i < len; ++i) {
int v = son[u][i];
if(v == ba) continue;
double tmp = dfs(v, u, r);
mi = min(mi, tmp);
dis1[u][v] = dis1[v][u] = min(dis1[v][u], tmp);
}
if(r != ba) mi = min(mi, ar[r][u]);
return mi;
}
LL two(LL x) {
return x * x;
}
void solve() {
scanf("%d%lf", &n, &m);
init();
for(int i = 1; i <= n; ++i) {
scanf("%lld%lld", &x[i], &y[i]);
}
for(int i = 1; i < n; ++i) {
for(int j = i + 1; j <= n; ++j) {
ar[i][j] = ar[j][i] = sqrt(two(x[i]-x[j])*1.0+two(y[i]-y[j]));
}
}
double tmp = prim(), ans = tmp;
//printf("MST = %.2f
", tmp);
for(int i = 1; i <= n; ++i) dfs(i, -1, i);
for(int i = 2; i < n; ++i) {
for(int j = i + 1; j <= n; ++j) {
if(is[i][j])ans = max(tmp - ar[i][j] + dis1[i][j], ans);
}
}
if(ans >= INF) printf("%.2f
", tmp * m);
else printf("%.2f
", ans * m);
}
int main(int argc, char const *argv[]){
int tim;scanf("%d", &tim);
while(tim--) solve();
return 0;
}
Problem Description：
一：
Our story is about Jebei Noyan(哲别), who was one of the most famous generals in Genghis Khan’s cavalry. Once his led the advance troop to invade a country named Pushtuar. The knights rolled up all the cities in Pushtuar rapidly. As Jebei Noyan’s advance troop did not have enough soldiers, the conquest was temporary and vulnerable and he was waiting for the Genghis Khan’s reinforce. At the meantime, Jebei Noyan needed to set up many guarders on the road of the country in order to guarantee that his troop in each city can send and receive messages safely and promptly through those roads.
There were N cities in Pushtuar and there were bidirectional roads connecting cities. If Jebei set up guarders on a road, it was totally safe to deliver messages between the two cities connected by the road. However setting up guarders on different road took different cost based on the distance, road condition and the residual armed power nearby. Jebei had known the cost of setting up guarders on each road. He wanted to guarantee that each two cities can safely deliver messages either directly or indirectly and the total cost was minimal.
Things will always get a little bit harder. As a sophisticated general, Jebei predicted that there would be one uprising happening in the country sooner or later which might increase the cost (setting up guarders) on exactly ONE road. Nevertheless he did not know which road would be affected, but only got the information of some suspicious road cost changes. We assumed that the probability of each suspicious case was the same. Since that after the uprising happened, the plan of guarder setting should be rearranged to achieve the minimal cost, Jebei Noyan wanted to know the new expected minimal total cost immediately based on current information.
Input
There are no more than 20 test cases in the input.
For each test case, the first line contains two integers N and M (1<=N<=3000, 0<=M<=N×N), demonstrating the number of cities and roads in Pushtuar. Cities are numbered from 0 to N-1. In the each of the following M lines, there are three integers xi, yi and ci(ci<=107), showing that there is a bidirectional road between xi and yi, while the cost of setting up guarders on this road is ci. We guarantee that the graph is connected. The total cost of the graph is less or equal to 109.
The next line contains an integer Q (1<=Q<=10000) representing the number of suspicious road cost changes. In the following Q lines, each line contains three integers Xi, Yi and Ci showing that the cost of road (Xi, Yi) may change to Ci (Ci<=107). We guarantee that the road always exists and Ci is larger than the original cost (we guarantee that there is at most one road connecting two cities directly). Please note that the probability of each suspicious road cost change is the same.
Output
For each test case, output a real number demonstrating the expected minimal total cost. The result should be rounded to 4 digits after decimal point.
Sample Input
3 3
0 1 3
0 2 2
1 2 5
3
0 2 3
1 2 6
0 1 6
0 0
Sample Output
6.0000
Hint
The initial minimal cost is 5 by connecting city 0 to 1 and city 0 to 2. In the first suspicious case, the minimal total cost is increased to 6;
the second case remains 5; the third case is increased to 7. As the result, the expected cost is (5+6+7)/3 = 6.
二：
As we all know, Nanjing is one of the four hottest cities in China. Students in NJUST find it hard to fall asleep during hot summer every year. They will never, however, suffer from that hot this year, which makes them really excited. NJUST’s 60th birthday is approaching, in the meantime, 50 million is spent to install air conditioning among students dormitories. Due to NJUST’s long history, the old circuits are not capable to carry heavy load, so it is necessary to set new high-load wires. To reduce cost, every wire between two dormitory is considered a segment. Now, known about all the location of dormitories and a power plant, and the cost of high-load wire per meter, Tom200 wants to know in advance, under the premise of all dormitories being able to supply electricity, the minimum cost be spent on high-load wires. And this is the minimum strategy. But Tom200 is informed that there are so many wires between two specific dormitories that we cannot set a new high-load wire between these two, otherwise it may have potential risks. The problem is that Tom200 doesn’t know exactly which two dormitories until the setting process is started. So according to the minimum strategy described above, how much cost at most you'll spend?
Input
The first line of the input contains a single integer T(T ≤ 100), the number of test cases.
For each case, the first line contains two integers n(3 ≤ n ≤ 1000), k(1 ≤ k ≤ 100). n represents n-1 dormitories and one power plant, k represents the cost of high-load wire per meter. n lines followed contains two integers x, y(0 ≤ x, y ≤ 10000000), representing the location of dormitory or power plant. Assume no two locations are the same, and no three locations are on a straight line. The first one is always the location of the power plant.
Output
For each case, output the cost, correct to two decimal places.
Sample Input
2
4 2
0 0
1 1
2 0
3 1
4 3
0 0
1 1
1 0
0 1
Sample Output
9.66
9.00