hdu 3790 双权值最短路径

来源：互联网发布：数据图的类型编辑：程序博客网时间：2024/06/05 09:12

只需要考虑最短路就好了，用dijkstra求最短路过程中只搜索了一条最短路，而本题只要在最短路相同时再比较花费谁最小就行了。对dijkstra做一点改动即可。

PS：有重边，输入时要做些处理。

代码如下：

#include <cstdio>#include <stack>#include <set>#include <iostream>#include <string>#include <vector>#include <queue>#include <list>#include <functional>#include <cstring>#include <algorithm>#include <cctype>#include <string>#include <map>#include <iomanip>#include <cmath>#define LL long long#define ULL unsigned long long#define SZ(x) (int)x.size()#define MP(a, b) make_pair(a, b)#define MS(arr, num) memset(arr, num, sizeof(arr))#define PB push_back#define F first#define S second#define ROP freopen("input.txt", "r", stdin);#define MID(a, b) (a + ((b - a) >> 1))#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define lrt rt << 1#define rrt rt << 1|1#define root 1,n,1#define BitCount(x) __builtin_popcount(x)#define BitCountll(x) __builtin_popcountll(x)#define LeftPos(x) 32 - __builtin_clz(x) - 1#define LeftPosll(x) 64 - __builtin_clzll(x) - 1const double PI = acos(-1.0);const int INF = 0x7fffffff;using namespace std;const double eps = 1e-5;const int MAXN = 300 + 10;const int MOD = 1000007;const int N=1100;const int MAX=log2(N*1.0);typedef pair<int, int> pii;typedef pair<int, string> pis;int n,m,g[N][N],val[N][N],d[N],value[N];bool vis[N];void dij(int s,int e){    int i,j;    for (i=1;i<=n;i++) value[i]= d[i]=INF;    d[s]=0;    value[s]=0;    MS(vis,false);    for (i=1;i<n;i++) {        int v,mn=INF;        for (j=1;j<=n;j++) if (!vis[j] && d[j]<mn) mn=d[v=j];        vis[v]=true;        for (j=1;j<=n;j++) if (!vis[j] && g[v][j]<INF) {            if ( d[j]>d[v]+g[v][j] || (d[j]==d[v]+g[v][j] && value[j]>value[v]+val[v][j]) ) { // 求最短路径或最短路径相同时花费最小的路径                d[j]=d[v]+g[v][j];                value[j]=value[v]+val[v][j];            }        }    }    printf("%d %d\n",d[e],value[e]);}int main(){    int i,j;    while(~scanf("%d%d",&n,&m),n)    {        for (i=1;i<=n;i++)            for (j=1;j<=n;j++) g[i][j]=val[i][j]=INF;        for (i=0;i<m;i++) {            int a,b,x,y;            scanf("%d%d%d%d",&a,&b,&x,&y);            if (x<g[a][b] || (x==g[a][b] && y<val[a][b]) ) {                g[a][b]=g[b][a]=x;                val[a][b]=val[b][a]=y;            }        }        int s,e;        scanf("%d%d",&s,&e);        dij(s,e);    }}

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