HDU 5886 - Tower Defence(最短路+最小割)

题目

题意：

n个点，m条边，构建有权无向图

告诉你有人从n点一定会选最短路走到1，然后你是1，要在某些地方设置障碍，要求断开N。

思路：

比赛时理解出了问题，以为是使1与n不连通，即求整个图的最小割，实际上题目中要求对方一定走最短路，是要先跑最短路，根据最短路来建图，在新图上跑最小割就可以了。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn = 1111,maxm = 11111;

struct Edge
{
int v,w,next;
} eg[maxm*5],teg[maxm*5];

int ds[maxn];
bool vis[maxn];
int head[maxn],thead[maxn];
int n,tp,ttp,st,en;

void Add(int u,int v,int w)
{
eg[tp].v = v;
eg[tp].w = w;
eg[tp].next = head[u];
head[u] = tp++;

eg[tp].v = u;
eg[tp].w = 0;
eg[tp].next = head[v];
head[v] = tp++;
}

void add(int u,int v,int w)
{
teg[ttp].v = v;
teg[ttp].w = w;
teg[ttp].next = thead[u];
thead[u] = ttp++;
}

int dis[maxn];
int gap[maxn];
int pre[maxn];
int cur[maxn];

int isap()
{
memset(dis,0,sizeof(dis));
memset(gap,0,sizeof(gap));
memset(pre,-1,sizeof(pre));
memset(cur,-1,sizeof(cur));

gap[0] = n;
int u,v,flow,ans;

flow = INF;
u = pre[0] = 0;
ans = 0;

while(dis[1] < en)
{
for(int &i = cur[u]; i != -1; i = eg[i].next)
{
if(eg[i].w && dis[eg[i].v]+1 == dis[u]) break;
}

if(cur[u] != -1)
{
v = eg[cur[u]].v;
flow = min(flow,eg[cur[u]].w);
pre[v] = u;
u = v;

if(u == en)
{
for(; u != 0; u = pre[u])
{
eg[cur[pre[u]]].w -= flow;
eg[cur[pre[u]]^1].w += flow;
}

ans += flow;
flow = INF;
}
}
else
{
int mn = n;
for(int i = head[u]; i != -1; i = eg[i].next)
{
if(eg[i].w && dis[eg[i].v] < mn)
{
mn = dis[eg[i].v];
cur[u] = i;
}
}

if(!(--gap[dis[u]])) break;
dis[u] = mn+1;
gap[dis[u]]++;
u = pre[u];
}
}
return ans;
}

void bfs()
{
queue <int> q;
memset(vis,0,sizeof(vis));
memset(ds,INF,sizeof(ds));
ds[0] = 0;
q.push(0);
int u,v;

while(!q.empty())
{
u = q.front();
q.pop();
vis[u] = 0;

for(int i = thead[u]; i != -1; i = teg[i].next)
{
v = teg[i].v;

if(ds[v] > ds[u]+1)
{
ds[v] = ds[u]+1;

if(!vis[v])
{
q.push(v);
vis[v] = 1;
}
}
}
}
}

int main()
{

int t;
int m;
int u,v,w;

scanf("%d",&t);

while(t--)
{
scanf("%d%d",&n,&m);

memset(head,-1,sizeof(head));
memset(thead,-1,sizeof(thead));
tp = ttp = 0;

st = 0,en = n-1;

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

u--,v--;
add(u,v,w);
add(v,u,w);
}

bfs();

for(int i = 0; i < n; ++i)
{
for(int j = thead[i]; j != -1; j = teg[j].next)
{
u = i;
v = teg[j].v;
if(ds[v] == ds[u]+1)
{
//printf("%d %d\n",u,v);
Add(u,v,teg[j].w);
}
}
}

printf("%d\n",isap());
}

return 0;
}

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