zoj 3229 Shoot the Bullet(无源汇上下界最大流)

题目：Shoot the Bullet

收藏：http://www.tuicool.com/articles/QRr2Qb

把每一天看成一个点，每个女孩也看成一个点，增加源和汇s、t，源向每一天连上[0,d]的边，每一天与每个女孩如果有拍照任务的话连上[l,r]的边，

每个女孩与汇连上[g,oo]的边，于是构成一个有上下界的图。

所以这道题目我们可以转换一下，只要连一条T → S的边，流量为无穷，没有下界，那么原图就得到一个无源汇的循环流图。

接下来的事情一样：原图中的边的流量设成自由流量ci – bi。新建源点SS汇点TT，求Mi，连边。然后求SS → TT最大流，判是否满流。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
typedef long long ll;
const int INF = 1<<30;
const int N = 1500;
const int M = 1e6;
int n, m, pre[N], d[N], cur[N], s, t, ss, tt;
int dis[N], in[N], low[36666], lcnt;
bool vis[N];

struct edge
{
    int u, v, cap, next;
    edge() {}
    edge(int u, int v, int c, int next):u(u),v(v),cap(c),next(next) {}
}e[M];
int ecnt;

void addedge(int u, int v, int w)
{
    e[ecnt] = edge(u, v, w, pre[u]);
    pre[u] = ecnt++;
    e[ecnt] = edge(v, u, 0, pre[v]);
    pre[v] = ecnt++;
}

bool BFS(int s, int t)
{
    memset(vis, 0, sizeof(vis));
    queue<int >q;
    q.push(s);
    vis[s] = 1;
    d[s] = 0;
    while(!q.empty())
    {
        int x = q.front(); q.pop();
        for(int i=pre[x]; ~i; i=e[i].next)
        {
            int v = e[i].v;
            if(!vis[v] && e[i].cap>0)
            {
                vis[v] = 1;
                d[v] = d[x] + 1;
                q.push(v);
            }
        }
    }
    return vis[t];
}

ll DFS(int x, ll c, int s, int t)
{
    if(x==t || c==0) return c;
    ll flow = 0, f;
    for(int &i = cur[x]; ~i; i = e[i].next)
    {
        int v = e[i].v;
        if(d[v] == d[x]+1 && (f=DFS(v, min(c, (ll)e[i].cap), s, t))>0)
        {
            e[i].cap -= f;
            e[i^1].cap += f;
            flow += f;
            c -= f;
            if(c==0) break;
        }
    }
    return flow;
}

ll Maxflow(int s, int t)
{
    ll flow = 0;
    while(BFS(s, t))
    {
        memcpy(cur, pre, sizeof(pre));
        flow += DFS(s, INF, s, t);
    }
    return flow;
}

int main()
{
//    freopen("in.txt", "r", stdin);
    while(scanf("%d%d", &n, &m)>0)
    {
        ll sum = 0;
        int g, id, R, k;
        s = 0, t = m+n+1;
        ss = n+m+2, tt = m+n+3;
        memset(pre, -1, sizeof(pre));
        memset(in, 0, sizeof(in));
        ecnt = 0, lcnt = 0;
        for(int i=1; i<=m; i++)
        {
            scanf("%d", &g);
            in[i+n] -= g;
            in[t] += g;
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d", &k, &dis[i]);
            for(int j=0; j<k; j++)
            {
                scanf("%d%d%d", &id, &low[++lcnt], &R);
                addedge(i, id+n+1, R-low[lcnt]);
                in[i] -= low[lcnt];
                in[id+n+1] += low[lcnt];
            }
        }
        for(int i= 1; i<=n; i++) addedge(s, i, dis[i]);
        for(int i= 1; i<=m; i++) addedge(i+n, t, INF);
        addedge(t, s, INF);
        for(int i = s; i<=t; i++)
        {
            if(in[i]>0) addedge(ss, i, in[i]), sum += in[i];
            if(in[i]<0) addedge(i, tt, -in[i]);
        }
        if(Maxflow(ss, tt)==sum)
        {
            Maxflow(s, t);
            addedge(t, s, -INF);
            int ans = 0;
            for(int i = pre[0]; ~i; i=e[i].next)
            {
                ans += e[i^1].cap ;
//                printf("u = %d, cap = %d
", e[i^1].u, e[i^1].cap);
            }
            printf("%d
", ans);
            for(int i=1; i<=lcnt; i++) printf("%d
", e[2*i-1].cap+low[i]);
        }
        else puts("-1");
        puts("");
    }
    return 0;
}