HDU 4940 - Destroy Transportation system(网络流）

题目：

http://acm.hdu.edu.cn/showproblem.php?pid=4940

题意：

给出一个强连通图，每条边有两个权值，破坏边的权值和修建边的权值。

询问是否存在一个集合S及其补集T，使得从S到T破换的边的权值和X，从T到S破坏的边和修建的边的权值和Y，满足X <= Y.

思路：

有上下界的网络流。

如果所有流出的流量等于流回的流量，那么便满足 X <= Y。

建图：

1：u到v 建一条 容量为 容量上届与下界差 的边。

2：s到u建 一条容量为 流入u的容量下界-流出u的容量下界 的边。若其容量w 为负数，则建 u到t 容量为 -w 的边。

AC.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
const int MAXN = 1000;
const int MAXM = 100010;
const int INF = 0x3f3f3f3f;

int w[MAXN];
struct Edge
{
int to, next, cap, flow;
} edge[MAXM];

int tol;
int head[MAXN];
int gap[MAXN], dep[MAXN], cur[MAXN];

void addedge(int u, int v, int w, int rw = 0)
{
edge[tol].to = v;
edge[tol].cap = w;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = rw;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}

int Q[MAXN];

void BFS(int start, int end)
{
memset(dep, -1, sizeof(dep));
memset(gap, 0, sizeof(gap));
gap[0] = 1;
int front = 0, rear = 0;
dep[end] = 0;
Q[rear++] = end;
while (front != rear)
{
int u = Q[front++];
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (dep[v] != -1)continue;
Q[rear++] = v;
dep[v] = dep[u] + 1;
gap[dep[v]]++;
}
}
}

int S[MAXN];

int sap(int start, int end, int N)
{
BFS(start, end);
memcpy(cur, head, sizeof(head));
int top = 0;
int u = start;
int ans = 0;
while (dep[start] < N)
{
if (u == end)
{
int Min = INF;
int inser;
for (int i = 0; i < top; i++)
if (Min > edge[S[i]].cap - edge[S[i]].flow)
{
Min = edge[S[i]].cap - edge[S[i]].flow;
inser = i;
}
for (int i = 0; i < top; i++)
{
edge[S[i]].flow += Min;
edge[S[i] ^ 1].flow -= Min;
}
ans += Min;
top = inser;
u = edge[S[top] ^ 1].to;
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] = i;
break;
}
}
if (flag)
{
S[top++] = cur[u];
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[S[--top] ^ 1].to;
}
return ans;
}
void init()
{
tol = 0;
memset(head, -1, sizeof(head));
memset(w, 0, sizeof(w));
}
int main()
{
//freopen("in", "r", stdin);
int T;
int ca = 1;
scanf("%d", &T);
while(T--) {
int n, m;
scanf("%d%d", &n, &m);
init();
int u, v, d, b;
for(int i = 0; i < m; ++i) {
scanf("%d%d %d%d", &u, &v, &d, &b);
addedge(u, v, b);
w[u] -= d;
w[v] += d;
}

int s = 0, t = n+1;
int sum = 0;
for(int i = 1; i <= n; ++i) {
if(w[i] > 0) {
sum += w[i];
addedge(s, i, w[i]);
}
if(w[i] < 0) {
addedge(i, t, -w[i]);
}
}

int flow = sap(s, t, n+2);
if(flow == sum) {
printf("Case #%d: happy\n", ca++);
}
else printf("Case #%d: unhappy\n", ca++);

}
return 0;
}