宁夏网络赛-F-Moving On
Posted 31415926535x
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https://www.cnblogs.com/31415926535x/p/11440395.html
一道简单的Floyd题,,但是是动态加点求多次有限制的最短路,,感觉这个思想很好,,当然可以直接dp
题意
题目给你一个图,然后对于每一个节点都有一个点权,然后有q次询问,每次询问两点间的最短距离,并且最短路径中不能通过任意一个点权大于等于w的点,(首尾不算),,
思路
一次询问的话,直接最短路乱搞就行了,,但是询问次数很多的时候,就不能每一次建图跑,因为是任意两点的最短路,而且给的图是邻接矩阵 ,所以用Floyed,,,但是怎么处理每一次的询问呢,,一种做法是再加一维,处理出任意的加入前k个点后的最短路,,最后回答询问即可,,,也就是dp的思想,,另一种是询问离线,动态建图跑q次floyed即可,,后面这种思路以前见过但是没套floyed用过,,
代码
离线
#include <bits/stdc++.h>
#define aaa cout<<233<<endl;
#define endl '\\n'
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
// mt19937 rnd(time(0));
const int inf = 0x3f3f3f3f;//1061109567 > 1e9
const ll linf = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-7;
const double pi = 3.14159265358979;
const int maxn = 2e2 + 5;
const int maxm = 2e4 + 5;
const int mod = 1e9 + 7;
int d[maxn][maxn];
struct query
int u, v, w;
int ans;
int id;
const bool operator<(const query &q)const
return w < q.w;
qry[maxm];
bool cmpid(query a, query b)
return a.id < b.id;
pair<int, int> r[maxn];
int main()
// double pp = clock();
// freopen("233.in", "r", stdin);
// freopen("233.out", "w", stdout);
ios_base::sync_with_stdio(0);
cin.tie(0);cout.tie(0);
int t; cin >> t;
int ca = 1;
while(t--)
int n, q; cin >> n >> q;
for(int i = 1; i <= n; ++i)cin >> r[i].first;
for(int i = 1; i <= n; ++i)r[i].second = i;
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
cin >> d[i][j];
for(int i = 1; i <= q; ++i)cin >> qry[i].u >> qry[i].v >> qry[i].w;
for(int i = 1; i <= q; ++i)qry[i].id = i;
sort(qry + 1, qry + 1 + q);
sort(r + 1, r + 1 + n);
int cnt = 1;
for(int qi = 1; qi <= q; ++qi)
while(cnt <= n && r[cnt].first <= qry[qi].w)
//满足条件的情况下,利用这个城市来更新最短路
int k = r[cnt].second;
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
++cnt;
qry[qi].ans = d[qry[qi].u][qry[qi].v];
sort(qry + 1, qry + 1 + q, cmpid);
cout << "Case #" << ca++ << ":" << endl;
for(int i = 1; i <= q; ++i)cout << qry[i].ans << endl;
// cout << endl << (clock() - pp) / CLOCKS_PER_SEC << endl;
return 0;
预处理在线
#include <bits/stdc++.h>
#define aaa cout<<233<<endl;
#define endl '\\n'
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
// mt19937 rnd(time(0));
const int inf = 0x3f3f3f3f;//1061109567 > 1e9
const ll linf = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-7;
const double pi = 3.14159265358979;
const int maxn = 2e2 + 5;
const int maxm = 2e4 + 5;
const int mod = 1e9 + 7;
int d[maxn][maxn][maxn];
pair<int, int> r[maxn];
int main()
// double pp = clock();
// freopen("233.in", "r", stdin);
// freopen("233.out", "w", stdout);
ios_base::sync_with_stdio(0);
cin.tie(0);cout.tie(0);
int t; cin >> t;
int ca = 1;
while(t--)
int n, q; cin >> n >> q;
for(int i = 1; i <= n; ++i)cin >> r[i].first;
for(int i = 1; i <= n; ++i)r[i].second = i;
memset(d, inf, sizeof d);
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
cin >> d[i][j][0];
sort(r + 1, r + 1 + n);
int cnt = 1;
for(int cnt = 1; cnt <= n; ++cnt)
int k = r[cnt].second;
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
d[i][j][cnt] = min(d[i][j][cnt - 1], d[i][k][cnt - 1] + d[k][j][cnt - 1]);
// sort(r + 1, r + 1 + n, [](pair<int, int> i, pair<int, int> j)return i.second < j.second;);
cout << "Case #" << ca++ << ":" << endl;
int u, v, w;
while(q--)
cin >> u >> v >> w;
int k = 0;
for(int i = 1; i <= n; ++i)if(r[i].first <= w)k = i;
cout << d[u][v][k] << endl;
// cout << endl << (clock() - pp) / CLOCKS_PER_SEC << endl;
return 0;
状态不在,思路都理不清,,,emmmmmm(该收心努力了啊314,,,,,
(end)
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