您的位置:首页 > 其它

ZOJ 3229 Shoot the Bullet

2012-08-14 21:50 281 查看
ZOJ_3229

有源汇有上下界的最大流问题,具体的思路可以参考这篇博客:http://blog.csdn.net/water_glass/article/details/6823741,感觉按套路来做就可以了。注意最后输出的时候不是要输出所有的MM。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 375
#define MAXD 1375
#define MAXM 225740
#define INF 0x3f3f3f3f
int N, M, first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM];
int S, T, SS, TT, d[MAXD], q[MAXD], work[MAXD], SUM;
struct List
{
int n, id[110];
}list[MAXD];
int low[MAXN][1010];
void add(int x, int y, int z)
{
v[e] = y, flow[e] = z;
next[e] = first[x], first[x] = e ++;
}
void init()
{
int i, j, x, G, D, C, high;
S = 0, T = N + M + 1, SS = T + 1, TT = SS + 1;
memset(first, -1, sizeof(first[0]) * (TT + 1)), e = 0;
SUM = 0;
for(i = 1; i <= M; i ++)
{
scanf("%d", &G);
add(N + i, T, INF - G), add(T, N + i, 0);
add(SS, T, G), add(T, SS, 0), add(N + i, TT, G), add(TT, N + i, 0);
SUM += G;
}
for(i = 1; i <= N; i ++)
{
scanf("%d%d", &C, &D), list[i].n = C;
add(S, i, D), add(i, S, 0);
for(j = 0; j < C; j ++)
{
scanf("%d", &x), ++ x;
list[i].id[j] = x;
scanf("%d%d", &low[i][x], &high);
add(i, N + x, high - low[i][x]), add(N + x, i, 0);
add(SS, N + x, low[i][x]), add(N + x, SS, 0), add(i, TT, low[i][x]), add(TT, i, 0);
SUM += low[i][x];
}
}
add(T, S, INF), add(S, T, 0);
}
int bfs(int S, int T)
{
int i, j, rear = 0;
memset(d, -1, sizeof(d[0]) * (TT + 1));
d[S] = 0, q[rear ++] = S;
for(i = 0; i < rear; i ++)
for(j = first[q[i]]; j != -1; j = next[j])
if(flow[j] && d[v[j]] == -1)
{
d[v[j]] = d[q[i]] + 1, q[rear ++] = v[j];
if(v[j] == T) return 1;
}
return 0;
}
int dfs(int cur, int a, int T)
{
if(cur == T) return a;
for(int &i = work[cur]; i != -1; i = next[i])
if(flow[i] && d[v[i]] == d[cur] + 1)
if(int t = dfs(v[i], std::min(a, flow[i]), T))
{
flow[i] -= t, flow[i ^ 1] += t;
return t;
}
return 0;
}
int dinic(int S, int T)
{
int ans = 0, t;
while(bfs(S, T))
{
memcpy(work, first, sizeof(first[0]) * (TT + 1));
while(t = dfs(S, INF, T))
ans += t;
}
return ans;
}
void print()
{
int i, j, k, ans = 0;
for(i = first[S]; i != -1; i = next[i]) if(v[i] != T) ans += flow[i ^ 1];
printf("%d\n", ans);
for(i = 1; i <= N; i ++)
{
for(j = first[i]; j != -1; j = next[j])
if(v[j] >= N + 1 && v[j] <= N + M)
low[i][v[j] - N] += flow[j ^ 1];
for(j = 0; j < list[i].n; j ++)
printf("%d\n", low[i][list[i].id[j]]);
}
}
void solve()
{
if(SUM != dinic(SS, TT))
printf("-1\n");
else
{
dinic(S, T);
print();
}
}
int main()
{
while(scanf("%d%d", &N, &M) == 2)
{
init();
solve();
printf("\n");
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: