传送门

题目大意

给出一个图,一些边带权,另一些边等待你赋权(最小赋为1).请你找到一种赋权方式,使得 s 到 t 的最短路为 L
n ≤ 1e3 ,m ≤ 1e4 ,L ≤ 1e9

分析

二分所有边的边权和

使得二分后第p条边权值为k,1~p-1条边权值为inf,剩余边权值为1

对于每种情况跑一次最短路

如果结果小于L则增大点权和否则减少

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
#define int long long
const int inf = 0x3f3f3f3f;
int n,m,s,t,L,d[10010],vis[10010];
priority_queue<pair<int,int> >q;
struct node {
    int x,y,z;
};
node a[10010];
int head[20020],w[20020],to[20020],nxt[20020],cnt;
vector<int>wh;
inline void add(int i){
    int x=a[i].x,y=a[i].y,z=a[i].z;
    nxt[++cnt]=head[x];
    head[x]=cnt;
    to[cnt]=y;
    w[cnt]=z;
    nxt[++cnt]=head[y];
    head[y]=cnt;
    to[cnt]=x;
    w[cnt]=z;
}
inline void dij(){
    d[s]=0;
    q.push(make_pair(0,s));
    while(!q.empty()){
      int x=q.top().second;
      q.pop();
      if(vis[x])continue;
      vis[x]=1;
      for(int i=head[x];i;i=nxt[i]){
          int y=to[i],z=w[i];
          if(d[y]>d[x]+z){
            d[y]=d[x]+z;
            q.push(make_pair(-d[y],y));
          }
      }
    }
}
inline int ck(int mid){
    int i,j,k;
    for(i=0;i<wh.size();i++){
      a[wh[i]].z=1+min(mid,inf);
      mid-=a[wh[i]].z-1;
    }
    memset(head,0,sizeof(head));
    memset(w,0,sizeof(w));
    memset(to,0,sizeof(to));
    memset(nxt,0,sizeof(nxt));
    cnt=0;
    for(i=1;i<=m;i++)add(i);
    memset(d,0x3f,sizeof(d));
    memset(vis,0,sizeof(vis));
    dij();
    return d[t];
}
signed main(){
    int i,j,k;
    scanf("%lld%lld%lld%lld%lld",&n,&m,&L,&s,&t);
    s++,t++;
    for(i=1;i<=m;i++){
      scanf("%lld%lld%lld",&a[i].x,&a[i].y,&a[i].z);
      a[i].x++,a[i].y++;
      if(!a[i].z)wh.push_back(i);
    }
    int le=0,ri=inf*wh.size();
    if(ck(le)>L||ck(ri)<L){
      puts("NO");
      return 0; 
    }
    puts("YES");
    while(ri-le>1){
      int mid=(le+ri)>>1;
      if(ck(mid)<=L)le=mid;
        else ri=mid;
    }
    ck(le);
    for(i=1;i<=m;i++)printf("%lld %lld %lld\n",a[i].x-1,a[i].y-1,a[i].z);
    return 0;
}