poj 3662 二分最短路（建模与转化）

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二分给出的p条边，把大于的都当做电信赠送的线路，小于等于的都当做农夫自己买的线路。

前者权值设为1，后者权值设为0，则最短路跑出来的结果和k值比较就知道是否满足，继续二分。

#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<vector>#include<algorithm>#include<set>#include<stack>#include<iterator>#include<queue>#include<ctime>#include<bitset>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll long long#define lson l,m,(rt<<1)using namespace std;#include<iostream>#include<math.h>#include<fstream>using namespace std;int n;int v[1024],path[1024];int map[1024][1024],dis[1024];int p,k;int Dijkstra(int s){     int i,j;     for(i=1;i<=n;i++)     {        v[i]=0;        dis[i]=map[s][i];     }     dis[s]=0;     for(i=1;i<=n;i++)     {        int min=10000000;        int pos=0;        for(j=1;j<=n;j++)        {            if(!v[j] )                if(dis[j]<min)                {                    pos=j;                    min=dis[j];                }        }        v[pos]=1;        for(j=1;j<=n;j++)        {            int p;            if(!v[j])                if(p=map[pos][j]+dis[pos],p<dis[j])                {                    dis[j]=p;                }        }    }    return dis[n];}struct node{    int x,y;    int d;}ve[200000];void build(int lim){    memset(map,0x3f,sizeof(map));    for(int i=1;i<=p;i++)    {        if(ve[i].d<=lim)        {            map[ve[i].x][ve[i].y]=map[ve[i].y][ve[i].x]=0;     //       cout<<v[i].x<<" "<<v[i].y<<endl;        }        else map[ve[i].x][ve[i].y]=map[ve[i].y][ve[i].x]=1;    }}bool cmp(node a,node b){    return a.d<b.d;}int main(){    int a,b,c;    while(scanf("%d%d%d",&n,&p,&k)!=EOF)    {        for(int i=1;i<=p;i++)        {            scanf("%d%d%d",&a,&b,&c);            ve[i].x=a;ve[i].y=b;ve[i].d=c;        }        sort(ve+1,ve+1+p,cmp);        build(0);        int val=Dijkstra(1);        if(val<=k)        {            printf("0\n");            continue;        }        else if(val>1000000)        {            printf("-1\n");            continue;        }        int l=1,r=p,mid;        int ans;        while(l<=r)        {            mid=(l+r)>>1;            build(ve[mid].d);            if(Dijkstra(1)<=k) {ans=mid;r=mid-1;}            else l=mid+1;        }        printf("%d\n",ve[ans].d);    }    return 0;}

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