HDU -- 1, 4370

1001. http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4370

A topic similar to God, which is modeled using graph theory ~ N has n points in total.

X<sub>12 + x13 +... x1n = 1
2. x1n + x2n +... xN-1n = 1</sub>

It is understood that the outbound degree of point 1 is 1, and the inbound degree of point N is 1;

The matrix is interpreted as the weight on the edge, so it can be understood:

1. The shortest path from 1 to n

2.1 and N respectively.

Mark wrote a dijikstra and orz with priority queue optimization, although I forgot to consider the self-ring.

#include<iostream>#include<cstdio>#include<queue>using namespace std;#define MAXN 330#define INF 100000000int n;int dis[MAXN];int g[MAXN][MAXN];bool vis[MAXN];priority_queue < pair<int,int> > pq;int dijkstra(){int i,S=0,T=n-1;vis[0]=dis[0]=0;pq.push(make_pair(0,S));for(i=1;i<n;i++) dis[i]=INF,vis[i]=false;while(!pq.empty()){int u=pq.top().second;pq.pop();if(vis[u]) continue;vis[u]=true;for(i=0;i<n;i++){if(dis[i]>dis[u]+g[u][i]){dis[i]=dis[u]+g[u][i];pq.push(make_pair(-dis[i],i));}}}return dis[T];}int main(){//freopen("data.in","r",stdin);//freopen("data.out","w",stdout);while(~scanf("%d",&n)){int i,j;for(i=0;i<n;i++)for(j=0;j<n;j++) scanf("%d",&g[i][j]);printf("%d\n",dijkstra());}return 0;}

Self-written dijisktra: a lot of frustration

# Include <cstdio> # include <iostream> using namespace STD; // complexity: O (N ^ 2) const int maxn = 330; const int INF = (1 <30); int mat [maxn] [maxn]; // s indicates the start point, D indicates the end point, and N indicates the matrix size, cannot process negative int Dijkstra (int s, int D, int N) {bool vis [maxn]; int dis [maxn], K = inf, one =-1; memset (VIS, false, sizeof (VIS); fill (DIS, DIS + N, INF); vis [s] = true; DIS [s] = 0; for (INT I = 0; I <n; I ++) dis [I] = mat [s] [I]; while (one! = D) {k = inf; For (INT I = 0; I <n; I ++) if (DIS [I] <K &&! Vis [I]) {one = I, K = dis [I];} vis [one] = true; // cout <"here" <one <Endl; for (INT I = 0; I <n; I ++) if (DIS [I]> dis [one] + mat [one] [I]) dis [I] = dis [one] + mat [one] [I];} return dis [d];} int main () {int N; while (~ Scanf ("% d", & N) {for (INT I = 0; I <n; I ++) for (Int J = 0; j <N; j ++) scanf ("% d", & mat [I] [J]); For (INT I = 0; I <n; I ++) mat [I] [I] = 0; printf ("% d \ n", Dijkstra (0, n-1, n);} return 0 ;}