[POJ 2391]Ombrophobic Bovines[最大流][二分答案]

题目链接：[POJ 2391]Ombrophobic Bovines[最大流][二分答案]
题意分析：

有F个地区，每个地区有have[i]头牛，可以提供can[i]头牛的庇护，有P条连接各个地区的无向道路，问：所有牛都能得到庇护，最少需要多少时间？

解题思路：

源点和牛间连一条容量为have[i]的边，汇点到牛间连一条容量为can[i]的边，需要将牛拆点（否则会发生串流，2->3 and 3->4 != 2->4）。用floyd预处理任意点间最短距离，二分答案，每次将距离小于等于答案的边添加进入图中，求最大流即可。如果满流就缩小答案，否则扩大答案。

个人感受：

第一想法是费用流，每次最大流的时候更新这条流的价值和到答案。然后写完WA了几发，自己给了自己一个反例：

3 3

2 2

0 2

2 0

1 2 10

3 1 10

3 2 100

ans: 10

然后floyd竟然写错了，太自信了= =。另外，网上一些过于优化的SAP也会WA，估计是优化过头出错误了。

具体代码如下：

#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<string>
#define lowbit(x) (x & (-x))
#define ll long long
#define pr(x) cout << #x << " = " << (x) << '\n';
using namespace std;

const int MAXN = 510;//点数的最大值
const int MAXM = 540010;//边数的最大值
const long long INF = 1e16;

struct Edge
{
int to,next,cap,flow;
}edge[MAXM];//注意是MAXM
int tol, src, des;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init()
{
tol = 0;
memset(head,-1,sizeof(head));
}
//加边，单向图三个参数，双向图四个参数
void addedge(int u,int v,int w,int rw=0)
{
edge[tol].to = v;edge[tol].cap = w;edge[tol].next = head[u];
edge[tol].flow = 0;head[u] = tol++;
edge[tol].to = u;edge[tol].cap = rw;edge[tol].next = head[v];
edge[tol].flow = 0;head[v]=tol++;
}
//输入参数：起点、终点、点的总数
//点的编号没有影响，只要输入点的总数
int sap(int start,int end,int N)
{
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,head,sizeof(head));
int u = start;
pre[u] = -1;
gap[0] = N;
int ans = 0;
while(dep[start] < N)
{
if(u == end)
{
int Min = INF;
for(int i = pre[u];i != -1; i = pre[edge[i^1].to])
if(Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
for(int i = pre[u];i != -1; i = pre[edge[i^1].to])
{
edge[i].flow += Min;
edge[i^1].flow -= Min;
}
u = start;
ans += Min;
continue;
}
bool flag = false;
int v;
for(int i = cur[u]; i != -1;i = edge[i].next)
{
v = edge[i].to;
if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
{
flag = true;
cur[u] = pre[v] = i;
break;
}
}
if(flag)
{
u = v;
continue;
}
int Min = N;
for(int i = head[u]; i != -1;i = edge[i].next)
if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
{
Min = dep[edge[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if(!gap[dep[u]])return ans;
dep[u] = Min+1;
gap[dep[u]]++;
if(u != start) u = edge[pre[u]^1].to;
}
return ans;
}

int have[MAXN], can[MAXN], f, sum;
ll mp[MAXN][MAXN];

void build(ll up) {
init();
for (int i = 1; i <= f; ++i) {
addedge(src, i, have[i]);
addedge(i + f, des, can[i]);
addedge(i, i + f, sum);
for (int j = i + 1; j <= f; ++j) {
if (mp[i][j] <= up) {
addedge(i, j + f, sum);
addedge(j, i + f, sum);
}
}
}
}

int main()
{
int m;
while (~scanf("%d%d", &f, &m)) {
sum = 0;
for (int i = 1; i <= f; ++i) {
scanf("%d%d", &have[i], &can[i]);
sum += have[i];
}
int u, v, w;
for (int i = 1; i <= f; ++i)
for (int j = 1; j <= f; ++j)
mp[i][j] = INF;

for (int i = 0; i < m; ++i) {
scanf("%d%d%d", &u, &v, &w);
if (w < mp[u][v])
mp[u][v] = mp[v][u] = w;
}

src = 0, des = 2 * f + 1;
for (int i = 1; i <= f; ++i) {
for (int j = 1; j <= f; ++j) {
for (int k = 1; k <= f; ++k)
mp[j][k] = min(mp[j][i] + mp[i][k], mp[j][k]);
}
}

ll ans = -1;
ll l = 1, r = INF - 1, mid;
while (l <= r) {
mid = (l + r) / 2;
build(mid);
if (sap(src, des, des + 1) >= sum) {
ans = mid;
r = mid - 1;
}
else l = mid + 1;
}
printf("%lld\n", ans);
}
return 0;
}