2012天津赛区网络赛第三题--Island Transport（hdu4280）

     题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4280

     题意：有N个岛屿之间有M双向条路，每条路每个小时最多能通过C个人，现在问一个小时内，最多能把多少个顾客从最西边的岛屿送至最东边的岛屿上。

     思路：网络流，求最大流。建图：每条路连接的两个岛屿之间建立一条容量为C的双向边，取超级源点与汇点，源点与最西边的岛屿，汇点与最东边的岛屿建立一条流量为无穷大的边。

      考点：网络流，Dinic非递归版，如果使用递归版的Dinic，将RE！

Code：

#include <stdio.h>
#include <string.h>
#define MAXN 100010
#define INF 0x3f3f3f
struct Edge{
int u;
int v;
int cap;
int next;
}edge[MAXN*2];
int head[MAXN];
int level[MAXN];
int que[MAXN];
int stack[MAXN];
int cur[MAXN];
int tot;
void addedge(int u,int v,int cap)
{
edge[tot].u = u;
edge[tot].v = v;
edge[tot].cap = cap;
edge[tot].next = head[u];
head[u] = tot;
tot++;

edge[tot].u = v;
edge[tot].v = u;
edge[tot].cap = cap;
edge[tot].next = head[v];
head[v] = tot;
tot++;
}
int BFS(int src,int des)
{
int i;
int front=0,real=0;
memset(level,-1,sizeof(level));
que[real++] = src;
level[src] = 0;
while(front != real)
{
int u = que[front++];
front = front%MAXN;
for(i = head[u];i != -1;i = edge[i].next)
{
int v = edge[i].v;
if(edge[i].cap > 0 && level[v] == -1)
{
level[v] = level[u] +1;
que[real++] = v;
real = real%MAXN;
if(v == des)
return 1;
}
}
}
return 0;
}
int dinic(int src,int des)
{
int i;
int res = 0;
while(BFS(src,des))
{
memcpy(cur,head,sizeof(head));
int u = src;
int top = 0;
while(1)
{
if(u == des)
{
int min = INF,id;
for(i=0;i<top;i++)
{
if(edge[stack[i]].cap < min)
{
min = edge[stack[i]].cap;
id = i;
}
}
for(i=0;i<top;i++)
{
edge[stack[i]].cap -= min;
edge[stack[i]^1].cap += min;
}
res += min;
top = id;
u = edge[stack[top]].u;
}

for(i=cur[u];i!=-1;i = cur[u] = edge[i].next)
if(edge[i].cap !=0 && level[u] == level[edge[i].v] -1)
break;

if(cur[u] != -1)
{
stack[top++] = cur[u];
u = edge[cur[u]].v;
}

else
{
if(top == 0)
break;
level[u] = -1;
u = edge[stack[--top]].u;
}
}
}
return res;
}
int main()
{
int i;
int ncase;
scanf("%d",&ncase);
while(ncase--)
{
int n,m;
int x,y,w;
int max = -INF,min = INF;
int idw,ide;
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
{
scanf("%d%d",&x,&y);
if(x>max)
{
max = x;
ide = i;
}
if(x<min)
{
min = x;
idw = i;
}
}
tot = 0;
memset(head,-1,sizeof(head));
addedge(0,idw,INF);
addedge(ide,n+1,INF);
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&x,&y,&w);
addedge(x,y,w);
}
int ans = dinic(0,n+1);
printf("%d\n",ans);
}
return 0;
}