POJ 2831：Can We Build This One?

Can We Build This One?

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 1415		Accepted: 527
Case Time Limit: 2000MS

Description

“Highways are built, then life is rich.” Now people of Big Town want to become rich, so they are planning to build highways to connect their villages.

Big Town is really big and has many villages. Its people plan to build some highways between some pairs of villages so that every pair of villages is connected by the highways either directly or indirectly. After surveying the geographical surroundings,
they find that there are some paths along with highways can be built. Every path is denoted by a triplet (a, b, c) which means a highway can built between the a-th village and the b-th village with a cost of c.
In order to save money, they will select only part of the paths to build highways along so that the total cost to build highways along the selected paths is minimal under the condition that every pair of villages is connected.

It is possible that multiple such selections exist. People from every village want to have those highways of good interest to them built. But some highways can never appear in the selection since they are much too costly. So people ask whether a certain
highway can be selected if they agree to cut the cost. Your task is to design a program to answer their queries.

Input

The first line of input contains three integers N, M and Q (1 < N ≤ 1,000, N − 1 ≤ M ≤ 100,000, 0 < Q ≤ 100,000), where N is the number of villages, M is the number of paths,
and Q is the number of queries. Each of the next M lines contains three integers a, b, and c (1 ≤ a, b ≤ N, a ≠ b, 0 ≤ c ≤ 1,000,000). The triplet (a, b, c)
describes a path. Each of following Q lines contains two integer i and x (1 ≤ i ≤ M, 0 ≤ x) describing a query, “Can a highway be built along the i-th path if the cost of is reduced to x?” x is
strictly lower than the original cost of building a highway along the i-th path. It is assumed that every pair of village will be connected either directly or indirectly if all possible highways are built. And there may be more than one highway that
can be built between a pair of villages.

Output

Output one line for each query. Output either “Yes” or “No” as the answer to the the query.

Sample Input

3 4 3
1 2 10
1 3 6
2 3 4
1 3 7
4 6
1 7
1 5

Sample Output

Yes
No
Yes

题意是询问一个边变小之后能不能进入到最小生成树的边中。

求最小生成树，计算两个点在最小生成树中的最大边，如果减过之后的边比这个最大边都小，那就可以替换到最小生成树中。

代码：

#pragma warning(disable:4996)
#include <iostream>
#include <functional>
#include <algorithm>
#include <cstring>
#include <vector>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <map>
using namespace std;
typedef long long ll;

#define INF 0x3fffffff
#define repp(i, n, m) for (int i = n; i <= m; i++)
#define rep(i, n, m) for (int i = n; i < m; i++)
#define sa(n) scanf("%d", &(n))
#define mp make_pair
#define ff first
#define ss second
#define pb push_back

const int maxn = 1e3 + 5;
const ll mod = 1000000;
const double PI = acos(-1.0);

int n, m, q;
int mini[maxn], vis[maxn], pre[maxn];
int dis[maxn][maxn], res[maxn][maxn];

struct ed
{
int u;
int v;
int w;
}edge[200005];

void init()
{
int i, j, k;
repp(i, 1, n)
{
repp(j, i, n)
{
if (i == j)
{
dis[i][j] = 0;
}
else
{
dis[i][j] = dis[j][i] = INF;
}
res[i][j] = res[j][i] = 0;
}
}
}

void prim()
{
int i, j, k, st;
int top = 0;
for (i = 1; i <= n; i++)
{
mini[i] = INF;
vis[i] = 0;
mini[i] = dis[1][i];
}
st = 1;
vis[1] = 1;
mini[1] = 0;

for (i = 1; i <= n - 1; i++)
{
for (j = 1; j <= n; j++)
{
if (vis[j] == 0 && dis[st][j] < mini[j])
{
mini[j] = dis[j][st];
pre[j] = st;
}
}
int min_all = INF;
for (j = 1; j <= n; j++)
{
if (vis[j] == 0 && mini[j] < min_all)
{
min_all = mini[j];
st = j;
}
}
for (j = 1; j <= n; j++)
{
if (vis[j])
{
res[j][st] = res[st][j] = max(res[j][pre[st]], mini[st]);
}
}
vis[st] = 1;
}
}

void solve()
{
int i, j, k;
int u, v, w;
repp(i, 1, m)
{
sa(edge[i].u), sa(edge[i].v), sa(edge[i].w);
dis[edge[i].u][edge[i].v] = dis[edge[i].v][edge[i].u] = min(dis[edge[i].v][edge[i].u], edge[i].w);
}
prim();
repp(i, 1, q)
{
sa(u), sa(w);
if (res[edge[u].u][edge[u].v] >= w)
{
puts("Yes");
}
else
{
puts("No");
}
}
}

int main()
{
while (scanf("%d%d%d", &n, &m, &q) != EOF)
{
init();
solve();
}
return 0;
}