标签:find void 大于 nal rest scanf blog done src
In Takahashi Kingdom, which once existed, there are N cities, and some pairs of cities are connected bidirectionally by roads. The following are known about the road network:
Snuke the archeologist found a table with N rows and N columns, A, in the ruin of Takahashi Kingdom. He thought that it represented the shortest distances between the cities along the roads in the kingdom.
Determine whether there exists a road network such that for each u and v, the integer Au,v at the u-th row and v-th column of A is equal to the length of the shortest path from City u to City v. If such a network exist, find the shortest possible total length of the roads.
相当于最小割 剪去每个不必要的边(在矩阵里是点) 每个点求到其他点的最短路 如果长度大于最短路则删去 如果等于就取节点最多的
#include<math.h> #include<string.h> #include<iostream> #include<stdio.h> #include<algorithm> #include<sstream> //istringstream stm(string); stm >> x; #include<vector> #include<queue> #define INF 2139062143 #define inf -2139062144 #define ll long long using namespace std; const int maxn = 1005; ll a[305][305]; ll tempa[305][305]; struct Edge { ll from, to, dist; Edge(ll u, ll v, ll d) : from(u), to(v), dist(d) {} }; struct HeapNode { ll d, u; bool operator<(const HeapNode &rhs) const { return d > rhs.d; } }; struct Dijkstra { ll n, m; vector <Edge> edges; vector<ll> G[maxn]; bool done[maxn]; ll d[maxn]; ll roadcount[maxn]; ll p[maxn]; //×??ì?·?Dμ?é?ò?ì??? vector<ll> path; void init(ll n) { this->n = n; for (ll i = 0; i < n; i++) G[i].clear(); edges.clear(); path.clear(); } void addEdge(ll from, ll to, ll dist) { edges.push_back(Edge(from, to, dist)); m = edges.size(); G[from].push_back(m - 1); } void dijkstra(ll s) { //′ósμ??aê? priority_queue <HeapNode> Q; for (ll i = 0; i <= n; i++) d[i] = INF; for (ll i = 0; i <= n; i++) roadcount[i] = 0; d[s] = 0; memset(done, 0, sizeof(done)); Q.push((HeapNode) { 0, s }); while (!Q.empty()) { HeapNode x = Q.top(); Q.pop(); ll u = x.u; if (done[u]) continue; done[u] = true; for (ll i = 0; i < G[u].size(); i++) { Edge &e = edges[G[u][i]]; if (d[e.to] > d[u] + e.dist) { d[e.to] = d[u] + e.dist; roadcount[e.to] = roadcount[u] + 1; p[e.to] = G[u][i]; Q.push((HeapNode) { d[e.to], e.to }); } else { if(d[e.to] == d[u] + e.dist) { roadcount[e.to] = max(roadcount[e.to],roadcount[u] + 1); } } } } } void findpath(ll x, ll end) { path.clear(); path.push_back(x); findd(x, end); } void findd(ll x, ll end) { if (x == end) return; else { path.push_back(this->edges[this->p[x]].from); findd(this->edges[this->p[x]].from, end); } } }; int main() { ll n; scanf("%lld",&n); Dijkstra dj; dj.init(n); for(ll i=0; i<n; i++) { for(ll j=0; j<n; j++) { scanf("%lld",&a[i][j]); tempa[i][j] = a[i][j]; if(i != j) dj.addEdge(i,j,a[i][j]); } } bool flag = true; for(ll i=0; i<n && flag; i++) { dj.dijkstra(i); for(ll j=0; j<n; j++) { if(a[i][j] > dj.d[j]) flag = false; if(dj.roadcount[j] > 1) { tempa[i][j] = 0; tempa[j][i] = 0; } } } if(!flag) printf("-1\n"); else { ll anss = 0; for(ll i=0; i<n; i++) { for(ll j=0; j<n; j++) { anss += tempa[i][j]; } } printf("%lld\n",anss / 2); } return 0; }
AtCoder Regular Contest 083 D:Restoring Road Network
标签:find void 大于 nal rest scanf blog done src
原文地址:http://www.cnblogs.com/Aragaki/p/7533161.html