To be an ACMan

dijkstra略微做下修改View Code #include<stdio.h>#include<string.h>#define maxn 1001#define INF 100000000int MIN(int a, int b){ return a < b ? a : b;}int dis[maxn],adj[maxn][maxn];bool vis[maxn];int n, m;void dijkstra(int v){ int i, j, k, u, max; for(i=1;i<=n;i++) { dis[i]=adj[v][i]; ... Read More

水题一枚数据规模不是很大，可以用floyd暴力过View Code #include<stdio.h>#include<string.h>#define INF 30#define maxn 101int dis[maxn][maxn];int n, m;void floyd(){ int i, j, k; for(k=1;k<=n;k++) for(i=1;i<=n;i++) for(j=1;j<=n;j++) if(k!=i&&k!=j) if(dis[i][k]+dis[k... Read More

HDU 1874 畅通工程续水dijkstraView Code #include<stdio.h>#include<string.h>#define INF 1000000#define maxn 201int adj[maxn][maxn],dis[maxn];//pre[maxn]bool vis[maxn];int n ,m;int dijkstra(int src, int dest){ int i ,j, k, u, min; for(i=0;i<n;i++) { vis[i]=0; dis[i]=adj[src][i]; ... Read More

题意：求源点1到其它任意点的最小值 之中的最大值。题意理解后，dijkstra水题View Code #include<stdio.h>#include<string.h>#define maxn 101#define INF 100000000int n, m;int adj[maxn][maxn],dis[maxn];int max;bool vis[maxn];int dijkstra(int v){ int i,j,k,min,u; for(i=1;i<=n;i++) { dis[i]=adj[v][i]; vis[i]=0; } ... Read More