SPOJ 371 Boxes

题意就是

有一些盒子，放在一个圈上，每个盒子中有若干个球，球的总数不会比盒子的数量多。

现在规定相邻的盒子之间可以把球移动过去，每次可以移动一个球，问用最少的步骤使得每个盒子中的球不超过1个

那么建图还是比较简单

源点跟每个点连接，容量为本来拥有的球数

每个点再与汇点连，容量为1

中间相邻的点之间连边，容量无穷，费用为1

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define eps 1e-5
#define MAXN 1111
#define MAXM 55555
#define INF 100000007
using namespace std;
struct EDGE
{
    int v, cap, cost, next, re;    //  re记录逆边的下标。
} edge[MAXM];
int n, m, ans, flow, src, des;
int e, head[MAXN];
int que[MAXN], pre[MAXN], dis[MAXN];
bool vis[MAXN];
void init()
{
    e = ans = flow = 0;
    memset(head, -1, sizeof(head));
}
void add(int u, int v, int cap, int cost)
{
    edge[e].v = v;
    edge[e].cap = cap;
    edge[e].cost = cost;
    edge[e].next = head[u];
    edge[e].re = e + 1;
    head[u] = e++;
    edge[e].v = u;
    edge[e].cap = 0;
    edge[e].cost = -cost;
    edge[e].next = head[v];
    edge[e].re = e - 1;
    head[v] = e++;
}
bool spfa()
{
    int i, h = 0, t = 1;
    for(i = 0; i <= n; i ++)
    {
        dis[i] = INF;
        vis[i] = false;
    }
    dis[src] = 0;
    que[0] = src;
    vis[src] = true;
    while(t != h)
    {
        int u = que[h++];
        h %= n;
        vis[u] = false;
        for(i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].v;
            if(edge[i].cap && dis[v] > dis[u] + edge[i].cost)
            {
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if(!vis[v])
                {
                    vis[v] = true;
                    que[t++] = v;
                    t %= n;
                }
            }
        }
    }
    if(dis[des] == INF) return false;
    return true;
}
void end()
{
    int u, p, mi = INF;
    for(u = des; u != src; u = edge[edge[p].re].v)
    {
        p = pre[u];
        mi = min(mi, edge[p].cap);
    }
    for(u = des; u != src; u = edge[edge[p].re].v)
    {
        p = pre[u];
        edge[p].cap -= mi;
        edge[edge[p].re].cap += mi;
        ans += mi * edge[p].cost;     //  cost记录的为单位流量费用，必须得乘以流量。
    }
    flow += mi;
}
int nt;
void build()
{
    int w;
    scanf("%d", &nt);
    init();
    src = nt + 1;
    des = nt + 2;
    n = des;
    for(int i = 1; i <= nt; i++)
    {
        scanf("%d", &w);
        add(src, i, w, 0);
        add(i, des, 1, 0);
    }
    for(int i = 1; i < nt; i++)
    {
        add(i, i + 1, INF, 1);
        add(i + 1, i, INF, 1);
    }
    add(1, nt, INF, 1);
    add(nt, 1, INF, 1);
}
void MCMF()
{
    init();
    build();
    while(spfa()) end();
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        MCMF();
        printf("%d\n", ans);
    }
    return 0;
}