hdu 2833 WuKong(最短路径+记忆化搜索)

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http://acm.hdu.edu.cn/showproblem.php?pid=2833

大致题意：给定一个无向图，以及悟空和师傅起点与终点，求它们分别从起点到终点的最短路径中经过相同的点的最大个数。

思路：首先dijkstra求出最短路，那么如果有dis[a] + map[a][b] = dis[b]，则边（a,b）一定在最短路径上。根据这一定理可以求出所有最短路径。然后类似于求最长公共子序列求经过的相同点的最大个数。

即若a==b ,dp[a][b] = max(dp[i][j]+1)

否则 dp[a][b] = max(dp[a][j],dp[i][b])，其中i,j分别是a,b的直接前驱节点。

#include <stdio.h>#include <iostream>#include <algorithm>#include <set>#include <map>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#define LL long long#define _LL __int64using namespace std;const int INF = 1<<28;const int maxn = 310;int Map[maxn][maxn];int n,m;int s1,t1,s2,t2;int dis1[maxn],dis2[maxn];int dp[maxn][maxn];void init(){    for(int i = 1; i <= n; i++)    {        for(int j = 1; j <= n; j++)            Map[i][j] = INF;    }}void dijkstra(int s, int f){    int dis[maxn],vis[maxn];    memset(vis,0,sizeof(vis));    for(int i = 1; i <= n; i++)        dis[i] = Map[s][i];    vis[s] = 1;    dis[s] = 0;    for(int i = 1; i <= n; i++)    {        int Min = INF,pos = -1;        for(int j = 1; j <= n; j++)        {            if(!vis[j] && dis[j] < Min)            {                Min = dis[j];                pos = j;            }        }        if(pos == -1) break;        vis[pos] = 1;        for(int j = 1; j <= n; j++)        {            if(!vis[j] && dis[j] > dis[pos] + Map[pos][j])                dis[j] = dis[pos] + Map[pos][j];        }    }    if(f == 1)        memcpy(dis1,dis,sizeof(dis));    else memcpy(dis2,dis,sizeof(dis));}int dfs (int a, int b){    if(a == s1 && b == s2)        return dp[s1][s2];    if(dp[a][b] > -1)        return dp[a][b]; //记忆化    int v = 0;    if(a == b)  // dp[a][b] = max(dp[i][j]+1)    {    v++;        for(int i = 1; i <= n; i++)        {            if(dis1[i] + Map[i][a] != dis1[a]) continue;            for(int j = 1; j <= n; j++)                if(dis2[j] + Map[j][b] == dis2[b])                    v = max(v,dfs(i,j)+1);        }        return dp[a][b] = v;    }//dp[a][b] = max(dp[i][b],dp[a][j]).    v = 0;    for(int i = 1; i <= n; i++){if(dis1[i] + Map[i][a] == dis1[a])v = max(v,dfs(i,b));}for(int i = 1; i <= n; i++){if(dis2[i] + Map[i][b] == dis2[b])v = max(v,dfs(a,i));}return dp[a][b] = v;}int main(){    while(~scanf("%d %d",&n,&m))    {        init();        if(n == 0 && m == 0) break;        int u,v,w;        for(int i = 1; i <= m; i++)        {            scanf("%d %d %d",&u,&v,&w);            if(w < Map[u][v])                Map[u][v] = Map[v][u] = w;        }        scanf("%d %d %d %d",&s1,&t1,&s2,&t2);        memset(dp,-1,sizeof(dp));        dp[s1][s2] = 0;        if(s1 == s2) // 注意起点相同的情况            dp[s1][s2] = 1;        dijkstra(s1,1);        dijkstra(s2,2);        int ans = dfs(t1,t2);        printf("%d\n",ans);    }    return 0;}

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