hdu 3416(最短路+最大流)

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题意：有 n 个城市，知道了起点和终点，有 m 条有向边，问从起点到终点的最短路一共有多少条。

解题思路：这题的关键就是找到哪些边可以构成最短路，其实之前做最短路的题目接触过很多，反向建一个图，求两边最短路，即从src到任一点的最短路dis1[]和从des到任一点的最短路dis2[]，那么假设这条边是(u,v,w)，如果dis1[u] + w + dis2[v] = dis1[des],说明这条边是构成最短路的边。找到这些边，就可以把边的容量设为1，跑一边最大流即可。

#include<iostream>#include<cstdio>#include<cstring>#include<queue>using namespace std;const int maxn = 1005;const int inf = 0x3f3f3f3f;struct Edge{int from,to,next,w;}edge[800005],E[100005];int n,m,st,ed;int cnt,head[maxn],pre[2][maxn];int dis[2][maxn],level[maxn];bool inq[maxn];void addedge(int u,int v,int w){edge[cnt].to = v;edge[cnt].w = w;edge[cnt].next = head[u];head[u] = cnt++;swap(u,v);edge[cnt].to = v;edge[cnt].w = 0;edge[cnt].next = head[u];head[u] = cnt++;}void addedge1(int u,int v,int w){edge[cnt].to = v;edge[cnt].w = w;edge[cnt].next = pre[0][u];pre[0][u] = cnt++;}void addedge2(int u,int v,int w){edge[cnt].to = v;edge[cnt].w = w;edge[cnt].next = pre[1][u];pre[1][u] = cnt++;}void build(){int u,v,w;memset(head,-1,sizeof(head));for(int i = 1; i <= m; i++){u = E[i].from, v = E[i].to, w = E[i].w;if(dis[0][u] + w + dis[1][v] == dis[0][ed])addedge(u,v,1);}}void spfa(int src,int des,int idx){queue<int> q;memset(dis[idx],inf,sizeof(dis[idx]));memset(inq,false,sizeof(inq));dis[idx][src] = 0;q.push(src);inq[src] = true;while(!q.empty()){int u = q.front();q.pop();inq[u] = false;for(int i = pre[idx][u]; i != -1; i = edge[i].next){int v = edge[i].to;if(dis[idx][v] > dis[idx][u] + edge[i].w){dis[idx][v] = dis[idx][u] + edge[i].w;if(inq[v] == false){inq[v] = true;q.push(v);}}}}}int BFS(int src,int des){    queue<int> q;    memset(level,0,sizeof(level));    level[src]=1;    q.push(src);    while(!q.empty()){        int u = q.front();        q.pop();        if(u==des) return 1;        for(int k = head[u];k!=-1;k=edge[k].next){            int v = edge[k].to,w=edge[k].w;            if(level[v]==0 && w!=0){                level[v]=level[u]+1;                q.push(v);            }        }    }    return -1;}int dfs(int u,int des,int increaseRoad){    if(u==des) return increaseRoad;    int ret=0;    for(int k=head[u];k!=-1;k=edge[k].next){        int v = edge[k].to,w=edge[k].w;        if(level[v]==level[u]+1&&w!=0){            int MIN = min(increaseRoad-ret,w);            w = dfs(v,des,MIN);if(w > 0){edge[k].w -=w;edge[k^1].w+=w;ret+=w;if(ret==increaseRoad) return ret;}else level[v] = -1;         }    }    return ret;}int Dinic(int src,int des){    int ans = 0;    while(BFS(src,des)!=-1) ans+=dfs(src,des,inf);    return ans;}int main(){int t,u,v,w;scanf("%d",&t);while(t--){cnt = 0;memset(pre,-1,sizeof(pre));scanf("%d%d",&n,&m);for(int i = 1; i <= m; i++){scanf("%d%d%d",&u,&v,&w);E[i].from = u,E[i].to = v,E[i].w = w;addedge1(u,v,w);addedge2(v,u,w);}scanf("%d%d",&st,&ed);spfa(st,ed,0);spfa(ed,st,1);build();int maxflow = Dinic(st,ed);printf("%d\n",maxflow);}return 0;}

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