poj 3228 Gold Transportation

                                                               Gold Transportation

题目链接：Click Here~

题目分析：

    For each case, output the minimum of the maximum adjacent distance on the condition that all the gold has been transported to the storehouses or "No Solution".

   题目理解没什么难得，主要是这一句的理解。翻译白话后其实就是：所有最短路中，最大的一条边。

算法分析：

    我们知道这道题中有多个源点和汇点。所以，我们要建立一个超级源点和超级汇点。然后，每次二分最长路。用最大流判断当前的最长路是否满足。其他细节自己看代码吧。

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 200+5;
const int INF = 10001;
struct Edge{
int from,to,cap,flow;
};
vector<Edge> edges;
vector<int> G[maxn];
bool vst[maxn];
int d[maxn];
int cur[maxn];
int gold[maxn],store[maxn],graph[maxn][maxn];
int scoure,sink,n;
void Init()
{
for(int i = 0;i <= n+1;++i)
G[i].clear();
edges.clear();
memset(d,0,sizeof(d));
}
void AddEdge(int from,int to,int cap)
{
edges.push_back((Edge){from,to,cap,0});
edges.push_back((Edge){to,from,0,0});
int sz = edges.size();
G[from].push_back(sz-2);
G[to].push_back(sz-1);
}
void Build(int num)
{
Init();
for(int i = 1;i <= n;++i)
for(int j = 1;j <= n;++j)
if(graph[i][j]<=num)
AddEdge(i,j,INF);
for(int i = 1;i <= n;++i)
AddEdge(scoure,i,gold[i]);
for(int i = 1;i <= n;++i)
AddEdge(i,sink,store[i]);
}
bool BFS() //xun zhao ceng ci tu
{
memset(vst,false,sizeof(vst));
queue<int> Q;
Q.push(scoure);
d[scoure] = 0;
vst[scoure] = 1;
while(!Q.empty()){
int x = Q.front();
Q.pop();
for(int i = 0;i < G[x].size();++i){
Edge& e = edges[G[x][i]];
if(!vst[e.to]&&e.cap>e.flow){
vst[e.to] = 1;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vst[sink];
}
int DFS(int x,int a)
{
if(x==sink||a==0)
return a;
int f,flow = 0;
for(int& i = cur[x];i < G[x].size();++i){
Edge& e = edges[G[x][i]];
if(d[x]+1==d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0){
e.flow += f;
edges[G[x][i]^1].flow -= f;
flow += f;
a -= f;
if(a==0)break;
}
}
return flow;
}
int Maxflow()
{
int flow = 0;
while(BFS()){
memset(cur,0,sizeof(cur));
flow += DFS(scoure,INF);
}
return flow;
}
int main()
{
while(scanf("%d",&n),n)
{
int sum = 0,tot = 0;
for(int i = 0;i <= n;++i)
for(int j = 0;j <= n;++j)
graph[i][j] = INF;
for(int i = 1;i <= n;++i){
scanf("%d",&gold[i]);
sum += gold[i];
}
for(int i = 1;i <= n;++i){
scanf("%d",&store[i]);
tot += store[i];
}
if(sum > tot){
puts("No Solution");
continue;
}
int m,x,y,d,maxv = -1,minv = INF;
scanf("%d",&m);
for(int i = 0;i < m;++i){
scanf("%d%d%d",&x,&y,&d);
graph[x][y] = graph[y][x] = d;
maxv = max(maxv,d);
minv = min(minv,d);
}
scoure = 0,sink = n+1;
int ans=-1,low = minv,high = maxv;
while(low <= high){
int mid = (low+high)>>1;
Build(mid);
int tmp = Maxflow();
if(tmp >= sum){
ans = mid;
high = mid -1;
}
else
low = mid + 1;
}
if(ans>0)
printf("%d\n",ans);
else
puts("No Solution");
}
return 0;
}